Chapter 5 Understanding Quadrilaterals & Practical Geometry

We need to identify the property that defines or describes a trapezium.

Let's examine each option:

(a) A pair of opposite sides is parallel.

This is the fundamental definition of a trapezium. A quadrilateral having at least one pair of parallel opposite sides is called a trapezium.

(b) The diagonals bisect each other.

This property is characteristic of parallelograms (including rhombuses, rectangles, and squares). A general trapezium does not have this property.

(c) The diagonals are perpendicular to each other.

This property is characteristic of rhombuses and kites. A general trapezium does not have this property.

(d) The diagonals are equal.

This property is characteristic of rectangles and isosceles trapeziums. A general trapezium does not necessarily have equal diagonals.

Therefore, the property that describes a trapezium is that it has a pair of opposite sides which are parallel.

The correct answer is (a) A pair of opposite sides is parallel.

We need to identify the property that is always true for any parallelogram.

Let's examine each option:

(a) Opposite sides are parallel.

This is the defining property of a parallelogram. By definition, a parallelogram is a quadrilateral with two pairs of parallel opposite sides.

(b) The diagonals bisect each other at right angles.

This property is true for a rhombus (a parallelogram with all sides equal) and a square (a parallelogram that is both a rhombus and a rectangle), but not for all parallelograms.

(c) The diagonals are perpendicular to each other.

This property is true for a rhombus and a square, but not for all parallelograms.

(d) All angles are equal.

This property means all angles are $90^\circ$. This is true for a rectangle (a parallelogram with all angles equal) and a square, but not for all parallelograms.

Therefore, the only property listed that is true for all parallelograms is that opposite sides are parallel.

The correct answer is (a) Opposite sides are parallel.

The sum of the interior angles of any quadrilateral is $360^\circ$.

An obtuse angle is an angle whose measure is greater than $90^\circ$ and less than $180^\circ$.

Let the four interior angles of the quadrilateral be $\alpha_1, \alpha_2, \alpha_3$, and $\alpha_4$.

We know that $\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 = 360^\circ$.

Suppose a quadrilateral has 4 obtuse angles.

Then, $\alpha_1 > 90^\circ$, $\alpha_2 > 90^\circ$, $\alpha_3 > 90^\circ$, and $\alpha_4 > 90^\circ$.

Summing these inequalities, we get:

$\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 > 90^\circ + 90^\circ + 90^\circ + 90^\circ = 360^\circ$.

This implies that the sum of the angles is greater than $360^\circ$, which contradicts the fact that the sum is exactly $360^\circ$.

Therefore, a quadrilateral cannot have 4 obtuse angles.

Suppose a quadrilateral has 3 obtuse angles.

Let $\alpha_1, \alpha_2, \alpha_3$ be obtuse angles, so $\alpha_1 > 90^\circ$, $\alpha_2 > 90^\circ$, $\alpha_3 > 90^\circ$.

The sum of these three angles is $\alpha_1 + \alpha_2 + \alpha_3 > 90^\circ + 90^\circ + 90^\circ = 270^\circ$.

The fourth angle is $\alpha_4 = 360^\circ - (\alpha_1 + \alpha_2 + \alpha_3)$.

Since $\alpha_1 + \alpha_2 + \alpha_3 > 270^\circ$, it follows that $\alpha_4 = 360^\circ - (\text{a value greater than } 270^\circ) < 360^\circ - 270^\circ = 90^\circ$.

So, the fourth angle $\alpha_4$ must be less than $90^\circ$ (i.e., it must be an acute angle or a right angle, but strictly less than $90^\circ$ if the other three are strictly greater than $90^\circ$).

We can construct a quadrilateral with 3 obtuse angles and 1 acute angle. For example, a concave quadrilateral can have angles $110^\circ, 110^\circ, 110^\circ$, and $30^\circ$. The sum is $110+110+110+30 = 360^\circ$. Here, three angles are obtuse.

Since a quadrilateral can have 3 obtuse angles, and cannot have 4, the maximum number of obtuse angles is 3.

The correct answer is (c) 3.

Consider a polygon with $n$ sides (an n-gon).

To divide the polygon into non-overlapping triangles by joining vertices, we can pick one vertex and draw diagonals from this vertex to all other non-adjacent vertices.

From a single vertex, we cannot draw a diagonal to itself or to its two adjacent vertices (as these form the sides of the polygon).

So, the number of vertices we can draw diagonals to from a chosen vertex is $n - 1 \text{ (the vertex itself)} - 2 \text{ (adjacent vertices)} = n - 3$.

Therefore, we can draw $n-3$ diagonals from one vertex of an n-gon.

These $n-3$ diagonals divide the polygon into triangles.

If we draw $n-3$ diagonals from one vertex, these diagonals, along with the sides of the polygon connected to that vertex, form triangles.

The number of triangles formed is always one more than the number of diagonals drawn from that single vertex.

Number of triangles $= \text{Number of diagonals} + 1 = (n-3) + 1 = n-2$.

These triangles are non-overlapping, and they completely cover the polygon.

Examples:

For a triangle ($n=3$), number of triangles $= 3-2=1$. (It's already a triangle).

For a quadrilateral ($n=4$), number of triangles $= 4-2=2$. (Drawing one diagonal divides it into 2 triangles).

For a pentagon ($n=5$), number of triangles $= 5-2=3$. (Drawing two diagonals from one vertex divides it into 3 triangles).

Thus, the maximum number of non-overlapping triangles we can make in an n-gon by joining vertices is $n-2$.

Comparing with the options:

(a) $n-1$

(b) $n-2$

(c) $n-3$

(d) $n-4$

The calculated number matches option (b).

The correct answer is (b) $n-2$.

A pentagon is a polygon with $n=5$ sides.

The sum of the interior angles of a polygon with $n$ sides is given by the formula:

Sum $= (n-2) \times 180^\circ$

For a pentagon, $n=5$. Substituting this value into the formula:

Sum $= (5-2) \times 180^\circ$

Sum $= 3 \times 180^\circ$

Sum $= 540^\circ$

Thus, the sum of all the interior angles of a pentagon is $540^\circ$.

Comparing this result with the given options:

(a) $180^\circ$

(b) $360^\circ$

(c) $540^\circ$

(d) $720^\circ$

The calculated sum matches option (c).

The correct answer is (c) 540°.

A hexagon is a polygon with $n=6$ sides.

The sum of the interior angles of a polygon with $n$ sides is given by the formula:

Sum $= (n-2) \times 180^\circ$

For a hexagon, $n=6$. Substituting this value into the formula:

Sum $= (6-2) \times 180^\circ$

Sum $= 4 \times 180^\circ$

Sum $= 720^\circ$

Thus, the sum of all the interior angles of a hexagon is $720^\circ$.

Comparing this result with the given options:

(a) $180^\circ$

(b) $360^\circ$

(c) $540^\circ$

(d) $720^\circ$

The calculated sum matches option (d).

The correct answer is (d) 720°.

In a parallelogram, adjacent angles are supplementary. This means their sum is $180^\circ$.

Let the two adjacent angles be $(5x - 5)^\circ$ and $(10x + 35)^\circ$.

Their sum is $180^\circ$.

$(5x - 5) + (10x + 35) = 180$

Combine like terms:

$5x + 10x - 5 + 35 = 180$

$15x + 30 = 180$

Subtract 30 from both sides:

$15x = 180 - 30$

$15x = 150$

Divide by 15 to solve for $x$:

$x = \frac{150}{15}$

$x = 10$

Now, calculate the measure of each angle by substituting $x=10$:

First angle: $5x - 5 = 5(10) - 5 = 50 - 5 = 45^\circ$

Second angle: $10x + 35 = 10(10) + 35 = 100 + 35 = 135^\circ$

The two adjacent angles are $45^\circ$ and $135^\circ$. Let's verify they are supplementary: $45^\circ + 135^\circ = 180^\circ$. This is correct.

Now, find the ratio of these angles:

Ratio $= 45^\circ : 135^\circ$

Divide both parts of the ratio by their greatest common divisor, which is 45.

$\frac{45}{45} = 1$

$\frac{135}{45} = 3$

The ratio is $1 : 3$.

Comparing the ratio with the given options, we find that the ratio $1 : 3$ matches option (a).

The correct answer is (a) 1 : 3.

Let's analyze the given properties of the quadrilateral:

1. All sides are equal: A quadrilateral with all sides equal is called a rhombus. A square also has all sides equal, so a square is a special type of rhombus.

2. Opposite angles are equal: This is a defining property of a parallelogram. Since a rhombus (and thus a square) is a type of parallelogram, it also satisfies this property.

3. The diagonals bisect each other at right angles: This is a property of a rhombus. The diagonals of a rhombus are perpendicular bisectors of each other. A square, being a rhombus, also has this property.

We are looking for a quadrilateral that satisfies all three of these properties.

Based on our analysis:

- Property 1 (All sides equal) narrows the possibilities to a rhombus or a square.

- Property 3 (Diagonals bisect each other at right angles) is a defining property of a rhombus. Any figure satisfying this property and having all sides equal must be a rhombus.

- Property 2 (Opposite angles are equal) is satisfied by any rhombus, as rhombuses are parallelograms.

Therefore, the quadrilateral that possesses all three given properties is a rhombus.

While a square also has these properties, the description perfectly matches the definition of a rhombus without requiring the additional property of having right angles (which distinguishes a square from other rhombuses).

Comparing this conclusion with the options:

(a) rhombus

(b) parallelogram

(c) square

(d) rectangle

The figure described is a rhombus.

The correct answer is (a) rhombus.

Let's analyze the given properties of the quadrilateral:

1. Opposite sides are equal: This is a property of a parallelogram. Rectangles, squares, and rhombuses are all types of parallelograms and thus have opposite sides equal.

2. All the angles are equal: A quadrilateral with all angles equal must have each angle equal to $\frac{360^\circ}{4} = 90^\circ$. Such a quadrilateral is a rectangle.

We are looking for a quadrilateral that satisfies both of these properties.

A quadrilateral with all angles equal ($90^\circ$) is a rectangle.

Rectangles also have opposite sides equal.

Therefore, a quadrilateral whose opposite sides and all the angles are equal is a rectangle.

Let's check the other options:

- A parallelogram has opposite sides equal, but not necessarily all angles equal (only opposite angles are equal).

- A square has all sides equal and all angles equal. Since all sides are equal, opposite sides are also equal. So, a square is a special type of rectangle. The description fits a square, but the option "rectangle" is a more general classification that also fits.

- A rhombus has all sides equal (so opposite sides are equal), but not necessarily all angles equal (only opposite angles are equal).

The description "opposite sides and all the angles are equal" perfectly matches the definition of a rectangle.

The correct answer is (a) rectangle.

We are given a quadrilateral with the following properties:

1. All sides are equal.

2. Diagonals are equal.

3. All angles are equal.

Let's analyze these properties:

Property 1 (All sides are equal) describes a rhombus. A square is a type of rhombus.

Property 3 (All angles are equal) implies that each angle must be $\frac{360^\circ}{4} = 90^\circ$. This describes a rectangle.

Property 2 (Diagonals are equal) describes a rectangle or an isosceles trapezium. However, combined with other properties, it helps specify the figure.

We are looking for a quadrilateral that is both a rhombus (all sides equal) and a rectangle (all angles equal, thus $90^\circ$).

A quadrilateral that is both a rhombus and a rectangle is a square.

Let's check if a square satisfies all three given properties:

- All sides are equal: Yes, this is a property of a square (it is a rhombus).

- Diagonals are equal: Yes, this is a property of a square (it is a rectangle).

- All angles are equal: Yes, all angles in a square are $90^\circ$ (it is a rectangle).

Thus, the quadrilateral that satisfies all three given properties is a square.

Comparing this conclusion with the given options:

(a) square

(b) trapezium

(c) rectangle

(d) rhombus

The figure described is a square.

The correct answer is (a) square.

A hexagon is a polygon with $n=6$ sides.

The number of diagonals $D$ in a polygon with $n$ sides is given by the formula:

$D = \frac{n(n-3)}{2}$

For a hexagon, we substitute $n=6$ into the formula:

$D = \frac{6(6-3)}{2}$

$D = \frac{6(3)}{2}$

$D = \frac{18}{2}$

$D = 9$

Thus, a hexagon has 9 diagonals.

Comparing this result with the given options:

(a) 9

(b) 8

(c) 2

(d) 6

The calculated number matches option (a).

The correct answer is (a) 9.

We are given a parallelogram with the property that its adjacent sides are equal.

Let the parallelogram be ABCD, with adjacent sides AB and BC being equal in length. So, AB = BC.

In a parallelogram, opposite sides are equal.

So, AB = CD and BC = AD.

Given AB = BC, and knowing the opposite side properties, we have:

AB = BC = CD = AD

This means that all four sides of the parallelogram are equal in length.

A parallelogram with all four sides equal is defined as a rhombus.

Let's consider the other options:

- A rectangle is a parallelogram with four right angles. Having equal adjacent sides does not guarantee right angles.

- A trapezium is a quadrilateral with only one pair of parallel sides. A parallelogram has two pairs, so it is not a trapezium.

- A square is a rhombus with four right angles. While a square fits the description, the property of having equal adjacent sides in a parallelogram only guarantees it is a rhombus, not necessarily a square (unless the angles are also $90^\circ$). The most accurate general term based solely on the given property is rhombus.

Thus, a parallelogram with equal adjacent sides is a rhombus.

The correct answer is (c) rhombus.

We are given a quadrilateral with two specific properties regarding its diagonals:

Property 1: The diagonals are equal.

Property 2: The diagonals bisect each other.

Let's consider the implications of these properties for different types of quadrilaterals.

If the diagonals of a quadrilateral bisect each other, this means the quadrilateral is a parallelogram.

So, the given quadrilateral is a parallelogram.

Now we also know that this parallelogram has equal diagonals.

Let's recall the properties of special parallelograms:

- A general parallelogram has diagonals that bisect each other, but they are not necessarily equal or perpendicular.

- A rhombus is a parallelogram where diagonals bisect each other at right angles. They are not necessarily equal in length (unless it's a square).

- A rectangle is a parallelogram where diagonals are equal in length and bisect each other. All angles are $90^\circ$.

- A square is a parallelogram where diagonals are equal in length, bisect each other, and are perpendicular. All sides are equal, and all angles are $90^\circ$.

The given properties are that the diagonals bisect each other (which makes it a parallelogram) and the diagonals are equal.

The parallelogram that has equal diagonals is a rectangle.

A square also satisfies these properties, but a rectangle is the general shape defined solely by being a parallelogram with equal diagonals.

Therefore, a quadrilateral whose diagonals are equal and bisect each other is a rectangle.

Comparing this conclusion with the given options:

(a) rhombus

(b) rectangle

(c) square

(d) parallelogram

The figure described is a rectangle.

The correct answer is (b) rectangle.

We need to find the sum of the exterior angles of a triangle, taking one exterior angle at each vertex.

A fundamental property of convex polygons states that the sum of the measures of the exterior angles, taking one at each vertex, is always $360^\circ$, regardless of the number of sides.

A triangle is a polygon with $n=3$ sides.

Applying the property mentioned above to a triangle, the sum of its exterior angles is $360^\circ$.

Alternatively, let the interior angles of the triangle be $\alpha, \beta, \gamma$. The sum of interior angles is $\alpha + \beta + \gamma = 180^\circ$.

Let the corresponding exterior angles be $\alpha', \beta', \gamma'$. Each interior angle and its adjacent exterior angle form a linear pair, so their sum is $180^\circ$.

$\alpha + \alpha' = 180^\circ \implies \alpha' = 180^\circ - \alpha$

$\beta + \beta' = 180^\circ \implies \beta' = 180^\circ - \beta$

$\gamma + \gamma' = 180^\circ \implies \gamma' = 180^\circ - \gamma$

Sum of exterior angles $= \alpha' + \beta' + \gamma' = (180^\circ - \alpha) + (180^\circ - \beta) + (180^\circ - \gamma)$

Sum $= 3 \times 180^\circ - (\alpha + \beta + \gamma)$

Substitute the sum of interior angles: $\alpha + \beta + \gamma = 180^\circ$.

Sum $= 540^\circ - 180^\circ$

Sum $= 360^\circ$

Both methods show that the sum of the exterior angles of a triangle is $360^\circ$.

Comparing this result with the given options:

(a) $180^\circ$

(b) $360^\circ$

(c) $540^\circ$

(d) $720^\circ$

The calculated sum matches option (b).

The correct answer is (b) 360°.

We are looking for a polygon that is both equiangular (all angles are equal) and equilateral (all sides are equal).

A polygon that is both equiangular and equilateral is called a regular polygon.

Let's examine the given options:

(a) Square: A square has four equal sides (equilateral) and four equal angles (all $90^\circ$, so equiangular). Thus, a square is a regular quadrilateral.

(b) Rectangle: A rectangle is equiangular (all angles are $90^\circ$) but not necessarily equilateral (adjacent sides can have different lengths). So, a rectangle is not a regular polygon unless it's a square.

(c) Rhombus: A rhombus is equilateral (all sides are equal) but not necessarily equiangular (opposite angles are equal, but adjacent angles are supplementary and generally not equal unless it's a square). So, a rhombus is not a regular polygon unless it's a square.

(d) Right triangle: A triangle has three sides and three angles. A right triangle has one angle equal to $90^\circ$. It is not generally equiangular (angles can be $30^\circ, 60^\circ, 90^\circ$ or other combinations) or equilateral (sides are related by the Pythagorean theorem and are generally not equal, except for an isosceles right triangle which is still not equilateral).

Among the given options, only a square is both equiangular and equilateral.

The correct answer is (a) Square.

We need to find a quadrilateral that possesses all the properties of both a kite and a parallelogram.

Let's list some key properties of a parallelogram:

- Opposite sides are parallel.

- Opposite sides are equal in length.

- Opposite angles are equal.

- Diagonals bisect each other.

Let's list some key properties of a kite:

- Two pairs of adjacent sides are equal in length.

- The diagonals are perpendicular to each other.

- One diagonal bisects the other diagonal.

- One pair of opposite angles (between unequal sides) are equal.

We are looking for a figure that is simultaneously a parallelogram and a kite.

If a figure is a parallelogram, its opposite sides are equal.

If this parallelogram is also a kite, it must have two pairs of congruent adjacent sides. Let the adjacent sides of the parallelogram be of lengths $a$ and $b$. Since opposite sides are equal, the side lengths are $a, b, a, b$. For it to have two pairs of congruent adjacent sides in the kite sense, we must have $a=b$. This means all four sides of the parallelogram are equal ($a, a, a, a$).

A parallelogram with all four sides equal is a rhombus.

Let's check if a rhombus satisfies the other properties of a kite:

- Does a rhombus have diagonals that are perpendicular? Yes, this is a property of a rhombus.

- Does a rhombus have one diagonal bisecting the other? Yes, this is a property of all parallelograms, and a rhombus is a parallelogram.

- Does a rhombus have two pairs of congruent adjacent sides? Yes, since all sides are equal, any two adjacent sides are congruent.

So, a rhombus has all the properties of a parallelogram and all the properties of a kite.

Let's examine the options:

(a) Trapezium: Is not a parallelogram.

(b) Rhombus: Is a parallelogram and also a kite (with all sides equal, the two pairs of equal adjacent sides are just formed by any two adjacent sides). Diagonals are perpendicular and bisect each other.

(c) Rectangle: Is a parallelogram, but generally not a kite (adjacent sides are not necessarily equal, diagonals are not perpendicular unless it's a square).

(d) Parallelogram: Does not necessarily have perpendicular diagonals or two pairs of equal adjacent sides unless it is a rhombus or square.

The figure that has all the properties of both a kite and a parallelogram is a rhombus.

The correct answer is (b) Rhombus.

The sum of the interior angles of a quadrilateral is $360^\circ$.

The angles of the quadrilateral are in the ratio $1 : 2 : 3 : 4$.

Let the angles be $x^\circ$, $2x^\circ$, $3x^\circ$, and $4x^\circ$, where $x$ is a constant.

According to the property of quadrilaterals, the sum of these angles is $360^\circ$.

$x + 2x + 3x + 4x = 360$

Combine the terms on the left side:

$(1 + 2 + 3 + 4)x = 360$

$10x = 360$

Solve for $x$:

$x = \frac{360}{10}$

$x = 36$

Now, calculate the measure of each angle using the value of $x=36$:

First angle: $x^\circ = 36^\circ$

Second angle: $2x^\circ = 2 \times 36^\circ = 72^\circ$

Third angle: $3x^\circ = 3 \times 36^\circ = 108^\circ$

Fourth angle: $4x^\circ = 4 \times 36^\circ = 144^\circ$

The four angles are $36^\circ, 72^\circ, 108^\circ, 144^\circ$.

Let's check if their sum is $360^\circ$:

$36 + 72 + 108 + 144 = 108 + 108 + 144 = 216 + 144 = 360^\circ$. This is correct.

The smallest angle is the one with the smallest multiple of $x$, which is $x^\circ$.

Smallest angle $= 36^\circ$.

Comparing this result with the given options:

(a) $72^\circ$

(b) $144^\circ$

(c) $36^\circ$

(d) $18^\circ$

The smallest angle matches option (c).

The correct answer is (c) 36°.

In the given trapezium ABCD, the arrows on sides AB and DC indicate that AB is parallel to DC.

When two parallel lines are intersected by a transversal, the consecutive interior angles are supplementary (their sum is $180^\circ$).

Here, AB || DC. The side AD acts as a transversal intersecting the parallel lines AB and DC.

Therefore, the angles $\angle A$ and $\angle D$ are consecutive interior angles.

We are given that $\angle A = 55^\circ$.

Substitute this value into the equation:

$55^\circ + \angle D = 180^\circ$

To find $\angle D$, subtract $55^\circ$ from $180^\circ$:

$\angle D = 180^\circ - 55^\circ$

$\angle D = 125^\circ$

Thus, the measure of $\angle D$ is $125^\circ$.

Note: The given values for $\angle B = 70^\circ$ and $\angle C = 115^\circ$ in the diagram are inconsistent with AB || DC, as $\angle B + \angle C$ should also be $180^\circ$ ($70^\circ + 115^\circ = 185^\circ \neq 180^\circ$). However, the question specifically asks for $\angle D$ and provides $\angle A$, and the parallel lines AB and DC imply the relationship between $\angle A$ and $\angle D$. We proceed with the calculation based on the property relating $\angle A$ and $\angle D$.

Comparing our result with the options:

(a) $55^\circ$

(b) $115^\circ$

(c) $135^\circ$

(d) $125^\circ$

Our calculated value matches option (d).

The correct answer is (d) 125°.

The sum of the interior angles of any quadrilateral is $360^\circ$.

We are given that three angles of the quadrilateral are acute, and each measures $80^\circ$.

An angle measuring $80^\circ$ is indeed acute since $0^\circ < 80^\circ < 90^\circ$.

Let the three given angles be $\alpha_1 = 80^\circ$, $\alpha_2 = 80^\circ$, and $\alpha_3 = 80^\circ$.

Let the fourth angle be $x^\circ$.

The sum of the four angles is $360^\circ$:

$\alpha_1 + \alpha_2 + \alpha_3 + x = 360^\circ$

$80^\circ + 80^\circ + 80^\circ + x = 360^\circ$

Sum the three known angles:

$240^\circ + x = 360^\circ$

Solve for $x$:

$x = 360^\circ - 240^\circ$

$x = 120^\circ$

The measure of the fourth angle is $120^\circ$.

Note that $120^\circ$ is an obtuse angle, which is possible for the remaining angle if three angles are acute.

Comparing our result with the given options:

(a) $150^\circ$

(b) $120^\circ$

(c) $105^\circ$

(d) $140^\circ$

Our calculated angle matches option (b).

The correct answer is (b) 120°.

For any convex polygon, the sum of the measures of its exterior angles (one at each vertex) is always $360^\circ$.

In a regular polygon with $n$ sides, all exterior angles are equal in measure.

If each exterior angle measures $45^\circ$, and there are $n$ such angles, their sum is $n \times 45^\circ$.

So, we have the equation:

$n \times 45^\circ = 360^\circ$

To find the number of sides $n$, divide $360^\circ$ by the measure of each exterior angle:

$n = \frac{360^\circ}{45^\circ}$

$n = \frac{360}{45}$

We can simplify the fraction:

$n = \frac{\cancel{360}^{72}}{\cancel{45}_{9}}$

$n = \frac{72}{9}$

$n = 8$

Thus, the regular polygon has 8 sides. A polygon with 8 sides is an octagon.

Comparing this result with the given options:

(a) 8

(b) 10

(c) 4

(d) 6

The calculated number of sides matches option (a).

The correct answer is (a) 8.

In a parallelogram, the following properties regarding angles hold:

1. Opposite angles are equal.

2. Adjacent angles are supplementary (sum up to $180^\circ$).

Given that in parallelogram PQRS, $\angle P = 60^\circ$.

The angle opposite to $\angle P$ is $\angle R$. Since opposite angles are equal:

$\angle R = \angle P = 60^\circ$

The angles adjacent to $\angle P$ are $\angle Q$ and $\angle S$. Since adjacent angles are supplementary:

$\angle P + \angle Q = 180^\circ$

$60^\circ + \angle Q = 180^\circ$

$\angle Q = 180^\circ - 60^\circ = 120^\circ$

Similarly, for the other adjacent angle $\angle S$:

$\angle P + \angle S = 180^\circ$

$60^\circ + \angle S = 180^\circ$

$\angle S = 180^\circ - 60^\circ = 120^\circ$

Also, the angles opposite to each other are equal, so $\angle Q$ should be equal to $\angle S$. We found $\angle Q = 120^\circ$ and $\angle S = 120^\circ$, which is consistent.

The four angles of the parallelogram are $\angle P = 60^\circ$, $\angle Q = 120^\circ$, $\angle R = 60^\circ$, and $\angle S = 120^\circ$.

The other three angles are $\angle Q, \angle R, \angle S$.

These are $120^\circ, 60^\circ, 120^\circ$. Listing them in numerical order or order of appearance (Q, R, S) gives $120^\circ, 60^\circ, 120^\circ$. Option (b) lists the values $60^\circ, 120^\circ, 120^\circ$, which are the values of the other three angles.

Comparing our calculated other three angles ($120^\circ, 60^\circ, 120^\circ$) with the options:

(a) $45^\circ, 135^\circ, 120^\circ$ (Incorrect)

(b) $60^\circ, 120^\circ, 120^\circ$ (Correct values, just potentially in a different order depending on which angle is listed first among the other three)

(c) $60^\circ, 135^\circ, 135^\circ$ (Incorrect)

(d) $45^\circ, 135^\circ, 135^\circ$ (Incorrect)

The other three angles are $60^\circ$ (opposite $\angle P$) and $120^\circ$ (adjacent to $\angle P$). The set of other three angles is $\{60^\circ, 120^\circ, 120^\circ\}$. Option (b) provides this set of values.

The correct answer is (b) 60°, 120°, 120°.

In a parallelogram, adjacent angles are supplementary. This means their sum is $180^\circ$.

The two adjacent angles are in the ratio $2 : 3$.

Let the measures of the adjacent angles be $2x^\circ$ and $3x^\circ$, where $x$ is a constant.

Since the angles are supplementary, their sum is $180^\circ$:

$2x + 3x = 180$

Combine the terms on the left side:

$5x = 180$

Solve for $x$:

$x = \frac{180}{5}$

$x = 36$

Now, calculate the measure of each adjacent angle by substituting $x=36$:

First angle: $2x^\circ = 2 \times 36^\circ = 72^\circ$

Second angle: $3x^\circ = 3 \times 36^\circ = 108^\circ$

The two adjacent angles are $72^\circ$ and $108^\circ$. Let's verify they are supplementary: $72^\circ + 108^\circ = 180^\circ$. This is correct.

Comparing these angles with the given options:

(a) $72^\circ, 108^\circ$

(b) $36^\circ, 54^\circ$

(c) $80^\circ, 120^\circ$

(d) $96^\circ, 144^\circ$

Our calculated angles match option (a).

The correct answer is (a) 72°, 108°.

In a parallelogram PQRS, the opposite angles are equal.

In parallelogram PQRS, $\angle P$ and $\angle R$ are opposite angles.

Therefore, $\angle P = \angle R$.

We are asked to find the value of $\angle P - \angle R$.

Since $\angle P = \angle R$, we can substitute $\angle P$ for $\angle R$ (or vice versa) in the expression $\angle P - \angle R$.

$\angle P - \angle R = \angle P - \angle P = 0^\circ$

Thus, $\angle P - \angle R = 0^\circ$.

Comparing our result with the given options:

(a) $60^\circ$

(b) $90^\circ$

(c) $80^\circ$

(d) $0^\circ$

Our calculated difference matches option (d).

The correct answer is (d) 0°.

In a parallelogram, adjacent angles are consecutive interior angles formed by a transversal intersecting a pair of parallel sides.

A key property of parallelograms is that consecutive (adjacent) interior angles are supplementary.

Supplementary angles are angles whose sum is $180^\circ$.

Therefore, the sum of adjacent angles of a parallelogram is always $180^\circ$.

For example, in parallelogram ABCD with AB || DC and AD || BC:

$\angle$A + $\angle$B = $180^\circ$

$\angle$B + $\angle$C = $180^\circ$

$\angle$C + $\angle$D = $180^\circ$

$\angle$D + $\angle$A = $180^\circ$

Comparing this property with the given options:

(a) $180^\circ$

(b) $120^\circ$

(c) $360^\circ$

(d) $90^\circ$

The sum of adjacent angles matches option (a).

The correct answer is (a) 180°.

In a parallelogram, the angle between the altitudes drawn from the vertex of the obtuse angle is equal to the measure of the adjacent acute angle.

Given that the angle between the altitudes is $30^\circ$, the measure of the acute angle of the parallelogram is $30^\circ$.

Adjacent angles in a parallelogram are supplementary (sum to $180^\circ$).

Let the obtuse angle be $\theta$.

$\theta + 30^\circ = 180^\circ$

Solving for $\theta$:

$\theta = 180^\circ - 30^\circ$

$\theta = 150^\circ$

The measure of the obtuse angle is $150^\circ$.

The correct answer is (b) 150°.

Given:

ABCD and BDCE are parallelograms.

Common base is DC.

BC $\perp$ BD.

To Find:

The measure of $\angle$BEC.

Solution:

Since BDCE is a parallelogram, the opposite sides are parallel and equal in length.

So, BD || CE and DC || BE.

Also, in parallelogram BDCE, the opposite angles are equal.

We are given that BC $\perp$ BD.

Since BD || CE (as established above), and BC is a line segment that intersects BD, BC acts as a transversal to the parallel lines BD and CE.

If a transversal is perpendicular to one of two parallel lines, it is perpendicular to the other as well.

Since BC $\perp$ BD and BD || CE, it follows that BC $\perp$ CE.

Now consider the triangles $\triangle$BCD and $\triangle$BCE.

In $\triangle$BCD, $\angle$CBD = $90^\circ$ (from given).

In $\triangle$BCE, $\angle$BCE = $90^\circ$ (from ii).

Let $\angle$BEC = $x$. From (i), $\angle$BDC = $x$.

In the right-angled triangle $\triangle$BCE (at C), we have:

In the right-angled triangle $\triangle$BCD (at B), we have:

From (iii) and (iv), we have:

Since BC is a side length, BC $\neq$ 0. We can divide both sides by BC.

This implies CE = BD, which we already know from the property of parallelogram BDCE. This comparison didn't help find $x$.

Let's use the other trigonometric ratio in $\triangle$BCD.

In the right-angled triangle $\triangle$BCD (at B),

The angles in $\triangle$BCD sum to $180^\circ$.

So, $\tan(90^\circ - x) = \frac{\text{BD}}{\text{BC}}$.

From (iv) and (vi), we have $\tan(x) = \frac{\text{BC}}{\text{BD}}$ and $\cot(x) = \frac{\text{BD}}{\text{BC}}$. This is consistent, as $\cot(x) = 1/\tan(x)$.

Let's look at $\triangle$BCE again.

In the right-angled triangle $\triangle$BCE (at C),

The angles in $\triangle$BCE sum to $180^\circ$.

So, $\tan(90^\circ - x) = \frac{\text{CE}}{\text{BC}}$.

From (vi) and (viii), we have:

Since BC $\neq$ 0, $\text{BD} = \text{CE}$. This is still the parallelogram property.

Let's try another approach using side ratios from both triangles with respect to BC.

From (iv): $\text{BC} = \text{BD} \tan(x)$.

From (viii): $\text{BC} = \text{CE} \cot(x)$.

Since BD = CE, we can write $\text{BC} = \text{BD} \cot(x)$.

So, $\text{BD} \tan(x) = \text{BD} \cot(x)$.

Since BD $\neq$ 0, we can divide by BD.

Since $x = \angle$BEC is an interior angle of a parallelogram, it must be positive. From the figure, $x = \angle$BDC appears acute, so $\tan(x)$ is positive.

The angle whose tangent is 1 is $45^\circ$.

Therefore, $\angle$BEC = $x = 45^\circ$.

Comparing this result with the given options:

(a) $60^\circ$

(b) $30^\circ$

(c) $45^\circ$

(d) $120^\circ$

Final Answer:

The measure of $\angle$BEC is $\mathbf{45^\circ}$.

The correct option is (c) $45^\circ$.

Given: Sides of a rectangle are 10 cm and 24 cm.

To Find: Length of one of the diagonals.

Solution:

In a rectangle, the diagonals are equal in length. Let the lengths of the sides of the rectangle be $l$ and $w$, and the length of the diagonal be $d$.

We can consider a right-angled triangle formed by two adjacent sides of the rectangle and a diagonal. The sides of the rectangle are the legs of the right-angled triangle, and the diagonal is the hypotenuse.

According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

So, $d^2 = l^2 + w^2$.

Given $l = 24$ cm and $w = 10$ cm.

$d^2 = (24)^2 + (10)^2$

$d^2 = 576 + 100$

$d^2 = 676$

To find the length of the diagonal $d$, we need to take the square root of 676.

$d = \sqrt{676}$

We can find the square root by prime factorization or by trial and error.

Using prime factorization:

$676 = 2 \times 2 \times 13 \times 13 = 2^2 \times 13^2 = (2 \times 13)^2 = 26^2$

So, $d = \sqrt{26^2} = 26$.

The length of the diagonal is 26 cm.

Comparing with the given options:

(a) 25 cm

(b) 20 cm

(c) 26 cm

(d) 3.5 cm

The correct option is (c).

The correct answer is (c) 26 cm.

Given: A parallelogram in which adjacent angles are equal.

To Find: The type of parallelogram.

Solution:

Let the parallelogram be ABCD. Let $\angle A$, $\angle B$, $\angle C$, and $\angle D$ be its interior angles.

In a parallelogram, adjacent angles are supplementary. This means the sum of any two adjacent angles is $180^\circ$.

So, $\angle A + \angle B = 180^\circ$, $\angle B + \angle C = 180^\circ$, $\angle C + \angle D = 180^\circ$, and $\angle D + \angle A = 180^\circ$.

We are given that adjacent angles are equal. Let's consider adjacent angles $\angle A$ and $\angle B$. We are given $\angle A = \angle B$.

Substitute this into the supplementary property equation:

$\angle A + \angle A = 180^\circ$

$2\angle A = 180^\circ$

$\angle A = \frac{180^\circ}{2}$

$\angle A = 90^\circ$

Since $\angle A = \angle B$, we have $\angle B = 90^\circ$.

In a parallelogram, opposite angles are equal. So, $\angle C = \angle A$ and $\angle D = \angle B$.

$\angle C = 90^\circ$ and $\angle D = 90^\circ$.

Thus, if the adjacent angles of a parallelogram are equal, all four angles are equal to $90^\circ$.

A parallelogram with all angles equal to $90^\circ$ is called a rectangle.

A rhombus is a parallelogram with all sides equal. Its angles are not necessarily $90^\circ$.

A trapezium is a quadrilateral with at least one pair of parallel sides; it is not necessarily a parallelogram.

Since the parallelogram has all angles equal to $90^\circ$, it is a rectangle.

Comparing with the given options:

(a) rectangle

(b) trapezium

(c) rhombus

(d) any of the three

The correct option is (a).

The correct answer is (a) rectangle.

Given: Four sets of angles.

To Find: Which set can be the interior angles of a quadrilateral.

Solution:

The sum of the interior angles of any quadrilateral is always equal to $360^\circ$.

To determine which set of angles can form a quadrilateral, we need to calculate the sum of the angles in each option and check if the sum is $360^\circ$.

Option (a): Angles are $140^\circ, 40^\circ, 20^\circ, 160^\circ$.

Sum $= 140^\circ + 40^\circ + 20^\circ + 160^\circ$

Sum $= (140 + 40) + (20 + 160)^\circ$

Sum $= 180^\circ + 180^\circ$

Sum $= 360^\circ$

The sum of the angles is $360^\circ$. All angles are positive and less than $180^\circ$, which is characteristic of a convex quadrilateral.

Option (b): Angles are $270^\circ, 150^\circ, 30^\circ, 20^\circ$.

Sum $= 270^\circ + 150^\circ + 30^\circ + 20^\circ$

Sum $= 420^\circ + 30^\circ + 20^\circ$

Sum $= 450^\circ + 20^\circ$

Sum $= 470^\circ$

The sum is not $360^\circ$.

Option (c): Angles are $40^\circ, 70^\circ, 90^\circ, 60^\circ$.

Sum $= 40^\circ + 70^\circ + 90^\circ + 60^\circ$

Sum $= 110^\circ + 90^\circ + 60^\circ$

Sum $= 200^\circ + 60^\circ$

Sum $= 260^\circ$

The sum is not $360^\circ$.

Option (d): Angles are $110^\circ, 40^\circ, 30^\circ, 180^\circ$.

Sum $= 110^\circ + 40^\circ + 30^\circ + 180^\circ$

Sum $= 150^\circ + 30^\circ + 180^\circ$

Sum $= 180^\circ + 180^\circ$

Sum $= 360^\circ$

The sum is $360^\circ$. However, one angle is $180^\circ$, which usually corresponds to a degenerate quadrilateral where three vertices are collinear.

Based on the sum property of interior angles of a quadrilateral, both option (a) and (d) result in a sum of $360^\circ$. However, option (a) represents a standard non-degenerate quadrilateral (specifically, a convex one), while option (d) represents a degenerate case.

Unless specified otherwise, questions about quadrilaterals typically refer to non-degenerate ones.

Therefore, the set of angles in option (a) can be the interior angles of a quadrilateral.

The correct answer is (a) 140°, 40°, 20°, 160°.

Given: A concave quadrilateral.

To Find: The sum of its interior angles.

Solution:

A concave quadrilateral (also known as a re-entrant quadrilateral) is a quadrilateral that has at least one interior angle greater than $180^\circ$. It has at least one diagonal that lies partly or entirely outside the quadrilateral.

The property that the sum of the interior angles of a quadrilateral is $360^\circ$ is true for any quadrilateral, whether it is convex or concave.

We can demonstrate this by dividing the quadrilateral into two triangles. In a concave quadrilateral, this can still be done using a diagonal. For example, if $\angle B$ is the reflex angle ($> 180^\circ$), we can draw the diagonal AC. This divides the quadrilateral ABCD into two triangles, $\triangle$ABC and $\triangle$ADC.

The sum of angles in $\triangle$ABC is $180^\circ$.

The sum of angles in $\triangle$ADC is $180^\circ$.

The sum of the interior angles of the quadrilateral is the sum of the angles of these two triangles, which is $180^\circ + 180^\circ = 360^\circ$.

Therefore, the sum of the interior angles of a concave quadrilateral is equal to $360^\circ$.

Comparing with the given options:

(a) more than $360^\circ$

(b) less than $360^\circ$

(c) equal to $360^\circ$

(d) twice of $360^\circ$ ($720^\circ$)

The correct option is (c).

The correct answer is (c) equal to 360°.

Given: Possible measures of the exterior angle of a regular polygon.

To Find: Which measure is not possible.

Solution:

For a regular polygon with $n$ sides, all exterior angles are equal.

The sum of the measures of the exterior angles of any convex polygon is always $360^\circ$.

Therefore, the measure of each exterior angle of a regular polygon with $n$ sides is given by the formula:

$\text{Exterior Angle} = \frac{360^\circ}{n}$

where $n$ is the number of sides of the polygon, and $n$ must be an integer greater than or equal to 3 (since a polygon must have at least 3 sides).

Conversely, if a given angle is the exterior angle of a regular polygon, then the number of sides $n = \frac{360^\circ}{\text{Exterior Angle}}$ must be an integer $\ge 3$.

We need to check which of the given options, when used in the formula $n = \frac{360^\circ}{\text{Angle}}$, does not result in an integer $\ge 3$.

Checking option (a): Angle $= 22^\circ$

$n = \frac{360}{22} = \frac{180}{11}$

The division $\frac{180}{11}$ does not result in an integer. 180 divided by 11 is approximately 16.36.

Since the number of sides $n$ is not an integer, $22^\circ$ cannot be the measure of an exterior angle of a regular polygon.

Checking option (b): Angle $= 36^\circ$

$n = \frac{360}{36} = 10$

The result $n=10$ is an integer and $10 \ge 3$. A regular decagon (10-sided polygon) has exterior angles of $36^\circ$. This is possible.

Checking option (c): Angle $= 45^\circ$

$n = \frac{360}{45} = 8$

The result $n=8$ is an integer and $8 \ge 3$. A regular octagon (8-sided polygon) has exterior angles of $45^\circ$. This is possible.

Checking option (d): Angle $= 30^\circ$

$n = \frac{360}{30} = 12$

The result $n=12$ is an integer and $12 \ge 3$. A regular dodecagon (12-sided polygon) has exterior angles of $30^\circ$. This is possible.

Only option (a) results in a non-integer number of sides. Therefore, $22^\circ$ can never be the measure of an exterior angle of a regular polygon.

The correct answer is (a) 22°.

Given: Four figures labeled (i), (ii), (iii), (iv).

To Find: Which figure represents a closed curve that is also a polygon.

Solution:

A closed curve is a curve that forms a loop, meaning it starts and ends at the same point.

A polygon is a simple closed curve made up entirely of line segments.

Let's examine each figure:

Figure (i): This figure is a circle. A circle is a closed curve, but it is made up of a continuous curve, not line segments. Therefore, it is not a polygon.

Figure (ii): This figure is a four-sided shape (a quadrilateral, specifically a square or rectangle based on appearance). It is a closed curve, and it is made up entirely of four straight line segments. Therefore, it is a polygon.

Figure (iii): This figure is a curve that does not form a closed loop. It has distinct start and end points. Therefore, it is not a closed curve and not a polygon.

Figure (iv): This figure is a closed curve. However, it contains a curved part (a semi-circle or arc). Since it is not made up entirely of line segments, it is not a polygon.

Based on the definitions, only Figure (ii) satisfies the condition of being both a closed curve and a polygon.

The options correspond to the figure labels:

(a) (ii)

(b) (iv)

(c) (i)

(d) (iii)

The figure that is a closed curve and also a polygon is (ii), which corresponds to option (a).

The correct answer is (a) (ii).

Given: Properties of a regular polygon with $n$ sides and its exterior angle.

To Find: Which given statement is not true.

Solution:

Let's analyse each statement based on the properties of regular polygons.

Statement (a): Each exterior angle = $\frac{360^\circ}{n}$

The sum of the exterior angles of any convex polygon (including a regular polygon) is $360^\circ$. In a regular polygon with $n$ sides, all $n$ exterior angles are equal. Therefore, the measure of each exterior angle is the sum of the exterior angles divided by the number of angles (which is equal to the number of sides $n$).

So, each exterior angle $= \frac{360^\circ}{n}$. This statement is true.

Statement (b): Exterior angle = $180^\circ$ – interior angle

At any vertex of a polygon, the interior angle and its corresponding exterior angle form a linear pair. Angles in a linear pair are supplementary, meaning their sum is $180^\circ$.

Interior angle + Exterior angle $= 180^\circ$

Rearranging this equation, we get:

Exterior angle $= 180^\circ$ – Interior angle

This statement is true.

Statement (c): $n = \frac{360^\circ}{extertor \ angle}$

Assuming "extertor angle" is a typo for "exterior angle", this statement is a rearrangement of the formula in statement (a). If we divide $360^\circ$ by the measure of each exterior angle, we get the number of sides $n$. This is valid for a regular polygon.

So, $n = \frac{360^\circ}{\text{Each exterior angle}}$. This statement is true.

Statement (d): Each exterior angle = $\frac{(n \;−\; 2) \;×\; 180°}{n}$

The sum of the interior angles of a polygon with $n$ sides is given by the formula $(n-2) \times 180^\circ$. In a regular polygon, all $n$ interior angles are equal. Therefore, the measure of each interior angle is the sum of interior angles divided by the number of angles $n$.

Each interior angle $= \frac{(n-2) \times 180^\circ}{n}$.

Comparing this with statement (d), we see that statement (d) gives the formula for the measure of an interior angle of a regular polygon, not the exterior angle.

Let's check if this interior angle formula equals the exterior angle formula using statement (b):

Exterior angle $= 180^\circ$ – Interior angle

Exterior angle $= 180^\circ - \frac{(n-2) \times 180^\circ}{n}$

Exterior angle $= 180^\circ \left( 1 - \frac{n-2}{n} \right)$

Exterior angle $= 180^\circ \left( \frac{n - (n-2)}{n} \right)$

Exterior angle $= 180^\circ \left( \frac{n - n + 2}{n} \right)$

Exterior angle $= 180^\circ \left( \frac{2}{n} \right) = \frac{360^\circ}{n}$.

This confirms that statement (d) provides the formula for the interior angle, not the exterior angle. Therefore, this statement is false for an exterior angle.

The statement that is not true for an exterior angle of a regular polygon with $n$ sides is (d).

The correct answer is (d) Each exterior angle = $\frac{(n \;−\; 2) \;×\; 180°}{n}$.

Given: PQRS is a square. Diagonals PR and SQ intersect at O.

To Find: The type of angle $\angle$POQ.

Solution:

A square is a special type of parallelogram where all sides are equal and all interior angles are $90^\circ$. It is also a rhombus (all sides equal) and a rectangle (all angles $90^\circ$).

One of the important properties of the diagonals of a square is that they:

1. Bisect each other.

2. Are equal in length.

3. Intersect at right angles.

The intersection point of the diagonals PR and SQ is O. The angle $\angle$POQ is one of the angles formed at the intersection of the diagonals.

According to the property of squares, the diagonals intersect at right angles.

Therefore, the angle formed by the intersection of the diagonals, $\angle$POQ, is $90^\circ$.

An angle that measures $90^\circ$ is called a right angle.

Comparing with the given options:

(a) Right angle ($90^\circ$)

(b) Straight angle ($180^\circ$)

(c) Reflex angle (greater than $180^\circ$ and less than $360^\circ$)

(d) Complete angle ($360^\circ$)

The correct option is (a).

The correct answer is (a) Right angle.

Given: Two adjacent angles of a parallelogram are in the ratio 1 : 5.

To Find: The measures of all four angles of the parallelogram.

Solution:

Let the two adjacent angles of the parallelogram be $\angle A$ and $\angle B$.

The ratio of the adjacent angles is given as 1 : 5.

Let the measures of the angles be $x$ and $5x$, where $x$ is a constant.

So, $\angle A = x$ and $\angle B = 5x$.

In a parallelogram, adjacent angles are supplementary. This means their sum is $180^\circ$.

Substitute the expressions for $\angle A$ and $\angle B$ into the equation:

$x + 5x = 180^\circ$

$6x = 180^\circ$

Divide both sides by 6:

$x = \frac{180^\circ}{6}$

$x = 30^\circ$

Now we can find the measures of the two adjacent angles:

$\angle A = x = 30^\circ$

$\angle B = 5x = 5 \times 30^\circ = 150^\circ$

So, two adjacent angles are $30^\circ$ and $150^\circ$.

In a parallelogram, opposite angles are equal. Let the angles be $\angle A, \angle B, \angle C, \angle D$ in order.

$\angle A = 30^\circ$

$\angle B = 150^\circ$

Opposite angle to $\angle A$ is $\angle C$, so $\angle C = \angle A = 30^\circ$.

Opposite angle to $\angle B$ is $\angle D$, so $\angle D = \angle B = 150^\circ$.

The four angles of the parallelogram are $30^\circ, 150^\circ, 30^\circ, 150^\circ$.

Let's check if these angles satisfy the properties of a parallelogram:

Adjacent angles sum to $180^\circ$: $30^\circ + 150^\circ = 180^\circ$. Correct.

Opposite angles are equal: $30^\circ = 30^\circ$ and $150^\circ = 150^\circ$. Correct.

Sum of all angles is $360^\circ$: $30^\circ + 150^\circ + 30^\circ + 150^\circ = 360^\circ$. Correct.

Comparing with the given options:

(a) $30^\circ, 150^\circ, 30^\circ, 150^\circ$

(b) $85^\circ, 95^\circ, 85^\circ, 95^\circ$

(c) $45^\circ, 135^\circ, 45^\circ, 135^\circ$

(d) $30^\circ, 180^\circ, 30^\circ, 180^\circ$ (Note: $180^\circ$ is not possible for an interior angle of a non-degenerate quadrilateral)

The correct option is (a).

The correct answer is (a) 30°, 150°, 30°, 150°.

Given: PQRS is a parallelogram with QR = 6 cm, PQ = 4 cm, and $\angle$PQR = $90^\circ$.

To Find: The type of parallelogram PQRS.

Solution:

PQRS is given to be a parallelogram.

In a parallelogram, opposite sides are equal in length.

Given PQ = 4 cm, so the opposite side RS = PQ = 4 cm.

Given QR = 6 cm, so the opposite side PS = QR = 6 cm.

The sides of the parallelogram are 4 cm, 6 cm, 4 cm, and 6 cm. Since adjacent sides PQ and QR have different lengths ($4 \text{ cm} \neq 6 \text{ cm}$), the parallelogram does not have all sides equal.

Given that one interior angle is $\angle$PQR = $90^\circ$.

In a parallelogram, adjacent angles are supplementary. So, $\angle$QPS + $\angle$PQR = $180^\circ$.

$\angle$QPS + $90^\circ = 180^\circ$

$\angle$QPS $= 180^\circ - 90^\circ = 90^\circ$.

Opposite angles of a parallelogram are equal. So, $\angle$PSR = $\angle$PQR = $90^\circ$ and $\angle$SRQ = $\angle$QPS = $90^\circ$.

Thus, all four angles of the parallelogram are $90^\circ$.

We have a parallelogram with:

1. Adjacent sides of unequal length (4 cm and 6 cm).

2. All interior angles equal to $90^\circ$.

Let's check the definitions of the given options:

(a) Square: A parallelogram with all sides equal and all angles equal to $90^\circ$. PQRS does not have all sides equal (4 cm and 6 cm). So, it is not a square.

(b) Rectangle: A parallelogram with one angle equal to $90^\circ$. Since PQRS is a parallelogram and $\angle$PQR = $90^\circ$, it fits the definition of a rectangle.

(c) Rhombus: A parallelogram with all sides equal. PQRS does not have all sides equal. So, it is not a rhombus.

(d) Trapezium: A quadrilateral with at least one pair of parallel sides. While a parallelogram is a type of trapezium, the specific properties given define it more precisely as a rectangle.

Since PQRS is a parallelogram with a right angle and unequal adjacent sides, it is a rectangle.

The correct answer is (b) rectangle.

Given: The angles P, Q, R, and S of a quadrilateral PQRS are in the ratio 1 : 3 : 7 : 9.

To Find: The type of quadrilateral PQRS.

Solution:

Let the angles of the quadrilateral be $\angle P, \angle Q, \angle R,$ and $\angle S$.

Their ratio is 1 : 3 : 7 : 9. Let the common ratio factor be $k$.

So, $\angle P = 1k$, $\angle Q = 3k$, $\angle R = 7k$, and $\angle S = 9k$.

The sum of the interior angles of a quadrilateral is $360^\circ$.

$\angle P + \angle Q + \angle R + \angle S = 360^\circ$

$k + 3k + 7k + 9k = 360^\circ$

Combine the terms:

$(1 + 3 + 7 + 9)k = 360^\circ$

$20k = 360^\circ$

Divide by 20:

$k = \frac{360^\circ}{20} = \frac{36}{2} = 18^\circ$

Now we can find the measure of each angle:

$\angle P = 1k = 1 \times 18^\circ = 18^\circ$

$\angle Q = 3k = 3 \times 18^\circ = 54^\circ$

$\angle R = 7k = 7 \times 18^\circ = 126^\circ$

$\angle S = 9k = 9 \times 18^\circ = 162^\circ$

The angles of the quadrilateral are $18^\circ, 54^\circ, 126^\circ, 162^\circ$.

Let's check the properties of different types of quadrilaterals:

(a) Parallelogram: Opposite angles are equal ($\angle P = \angle R$, $\angle Q = \angle S$) and adjacent angles are supplementary ($\angle P + \angle Q = 180^\circ$, etc.).

$\angle P = 18^\circ$, $\angle R = 126^\circ$. $\angle P \neq \angle R$. Not a parallelogram.

(b) Trapezium with PQ || RS: If PQ || RS, then consecutive interior angles are supplementary. This means $\angle P + \angle S = 180^\circ$ and $\angle Q + \angle R = 180^\circ$.

$\angle P + \angle S = 18^\circ + 162^\circ = 180^\circ$.

$\angle Q + \angle R = 54^\circ + 126^\circ = 180^\circ$.

Since $\angle P + \angle S = 180^\circ$ and $\angle Q + \angle R = 180^\circ$, the lines PS and QR are transversal to lines PQ and RS. The sum of interior angles on the same side of the transversal is $180^\circ$, which implies PQ || RS.

Since it has at least one pair of parallel sides (PQ || RS), it is a trapezium. The condition $\angle P + \angle S = 180^\circ$ and $\angle Q + \angle R = 180^\circ$ confirms that PQ || RS.

(c) Trapezium with QR || PS: If QR || PS, then consecutive interior angles are supplementary. This means $\angle P + \angle Q = 180^\circ$ and $\angle R + \angle S = 180^\circ$.

$\angle P + \angle Q = 18^\circ + 54^\circ = 72^\circ \neq 180^\circ$.

So, QR is not parallel to PS.

(d) Kite: A kite has two pairs of equal-length adjacent sides. Its angles can vary. One property is that one pair of opposite angles are equal, and the diagonal between the unequal sides bisects the other two angles and is perpendicular to the other diagonal. The angle measures $18^\circ, 54^\circ, 126^\circ, 162^\circ$ do not suggest a kite unless there is a specific relationship between these angles that defines a kite. The sum of the unequal opposite angles (say, $54^\circ$ and $162^\circ$) is $216^\circ$, and the sum of the equal opposite angles (if any) must be $360^\circ - 216^\circ = 144^\circ$, meaning each equal angle would be $72^\circ$. This does not match the angles $18^\circ$ and $126^\circ$. So, it is not a kite based on angle properties.

Based on the angle calculations, the quadrilateral is a trapezium with PQ || RS.

The correct answer is (b) trapezium with PQ || RS.

Given: PQRS is a trapezium with PQ || SR, $\angle$P = $130^\circ$, and $\angle$Q = $110^\circ$.

To Find: The measure of $\angle$R.

Solution:

In a trapezium, the sum of consecutive interior angles between the parallel sides is $180^\circ$.

Given that PQ || SR.

The angles $\angle$P and $\angle$S are consecutive interior angles between the parallel lines PQ and SR, with PS as the transversal.

The angles $\angle$Q and $\angle$R are consecutive interior angles between the parallel lines PQ and SR, with QR as the transversal.

Therefore, we have the following relationships:

We are given $\angle$Q = $110^\circ$. We need to find $\angle$R.

Using the second equation:

$110^\circ + \angle$R $= 180^\circ$

Subtract $110^\circ$ from both sides:

$\angle$R $= 180^\circ - 110^\circ$

$\angle$R $= 70^\circ$

We can also find $\angle$S using the first equation and the given $\angle$P = $130^\circ$, although it is not required by the question:

$130^\circ + \angle$S $= 180^\circ$

$\angle$S $= 180^\circ - 130^\circ = 50^\circ$.

The angles of the trapezium are $\angle$P = $130^\circ$, $\angle$Q = $110^\circ$, $\angle$R = $70^\circ$, $\angle$S = $50^\circ$.

Let's check the sum of angles: $130^\circ + 110^\circ + 70^\circ + 50^\circ = 240^\circ + 120^\circ = 360^\circ$. The sum is correct for a quadrilateral.

The question asks for the value of $\angle$R.

Comparing with the given options:

(a) $70^\circ$

(b) $50^\circ$

(c) $65^\circ$

(d) $55^\circ$

The correct option is (a).

The correct answer is (a) 70°.

Given: Each interior angle of a regular polygon = $135^\circ$.

To Find: Number of sides ($n$).

Solution:

The exterior angle of a regular polygon is supplementary to its interior angle.

Exterior angle $= 180^\circ$ – Interior angle

Exterior angle $= 180^\circ - 135^\circ = 45^\circ$.

For a regular polygon with $n$ sides, each exterior angle is $\frac{360^\circ}{n}$.

Solving for $n$:

$n = \frac{360^\circ}{45^\circ}$

$n = 8$

The number of sides is 8.

Comparing with the given options, the correct option is (c).

The correct answer is (c) 8.

Given: A quadrilateral where a diagonal bisects both the angles at its endpoints.

To Find: The type of quadrilateral.

Solution:

Let the quadrilateral be ABCD and let the diagonal AC bisect angles $\angle$A and $\angle$C.

This means that $\angle$BAC = $\angle$DAC and $\angle$BCA = $\angle$DCA.

Consider triangles $\triangle$ABC and $\triangle$ADC.

We have:

AC = AC (Common side)

$\angle$BAC = $\angle$DAC (Given that AC bisects $\angle$A)

$\angle$BCA = $\angle$DCA (Given that AC bisects $\angle$C)

By the Angle-Side-Angle (ASA) congruence criterion, if two angles and the included side of one triangle are equal to two angles and the included side of another triangle, then the triangles are congruent. However, here the side AC is included between $\angle$BAC and $\angle$BCA in $\triangle$ABC, and between $\angle$DAC and $\angle$DCA in $\triangle$ADC.

So, by ASA congruence:

$\triangle$ABC $\cong$ $\triangle$ADC

By Corresponding Parts of Congruent Triangles (CPCTC), the corresponding sides are equal.

AB = AD

CB = CD

This property, where two pairs of adjacent sides are equal (AB=AD and CB=CD), is the definition of a kite.

In a kite, the diagonal between the vertices where the equal sides meet (vertices A and C in this case, where AB=AD and CB=CD) is the axis of symmetry. This diagonal bisects the angles at these two vertices ($\angle$A and $\angle$C) and is perpendicular to the other diagonal (BD).

Let's consider the given options:

(a) Kite: A kite has two pairs of equal-length adjacent sides. As shown above, the given property implies that the quadrilateral has two pairs of equal adjacent sides, which is the definition of a kite. Also, the diagonal connecting the vertices between the equal sides bisects the angles at these vertices. This matches the property.

(b) Parallelogram: In a general parallelogram, the diagonals do not bisect the angles. This property only holds for a rhombus (or square), which is a special type of parallelogram.

(c) Rhombus: A rhombus is a parallelogram with all four sides equal. A rhombus is a special case of a kite where all four sides are equal (so AB=AD=CB=CD). In a rhombus, *both* diagonals bisect the angles they connect. So, if a quadrilateral is a rhombus, a diagonal (any diagonal) does bisect the angles. However, the property (a diagonal bisects angles) defines a kite. A kite is not necessarily a rhombus.

(d) Rectangle: In a general rectangle, the diagonals do not bisect the angles unless it is a square.

The property "a diagonal of a quadrilateral bisects both the angles [at its endpoints]" is the defining characteristic of a kite.

The correct answer is (a) kite.

Given: The requirement to construct a unique parallelogram.

To Find: The minimum number of measurements needed for unique construction.

Solution:

To uniquely construct a parallelogram, we need enough information to fix its shape and size.

Common ways to uniquely define a parallelogram include:

• Lengths of two adjacent sides and the measure of the angle between them (e.g., side-angle-side).
• Lengths of the two diagonals and the measure of the angle between them.
• Length of one side and the lengths of both diagonals.

Each of these methods requires 3 measurements.

With only 2 measurements (e.g., two adjacent side lengths, or one side and one angle), the parallelogram is not unique; its angles or other dimensions can vary.

Therefore, the minimum number of measurements required to construct a unique parallelogram is 3.

The correct answer is (b) 3.

Given: The requirement to construct a unique rectangle.

To Find: The minimum number of measurements needed for unique construction.

Solution:

A rectangle is a special type of parallelogram in which all interior angles are equal to $90^\circ$. Since the angles are fixed at $90^\circ$, we do not need angle measurements to define the shape itself.

A rectangle is uniquely determined by the lengths of its adjacent sides, commonly referred to as its length and width.

Let the length of the rectangle be $l$ and the width be $w$. If we are given the values of $l$ and $w$, we can construct the rectangle as follows:

1. Draw a line segment AB of length $l$.

2. At point A, draw a ray perpendicular to AB.

3. On this perpendicular ray, mark a point D such that AD $= w$.

4. From point B, draw a ray perpendicular to AB (or parallel to AD).

5. From point D, draw a ray parallel to AB (or perpendicular to AD).

6. The intersection of the rays from B and D is point C. ABCD is the rectangle.

This construction uniquely determines the rectangle using only two measurements: the length ($l$) and the width ($w$).

Alternatively, a unique rectangle can also be constructed if we are given the length of one side and the length of a diagonal. Let the side be $l$ and the diagonal be $d$. Since the angle between the side and the adjacent side is $90^\circ$, we can form a right-angled triangle with sides $l$, the unknown width $w$, and the hypotenuse $d$. By the Pythagorean theorem, $w^2 = d^2 - l^2$. If $d > l$, $w = \sqrt{d^2 - l^2}$, which is a unique positive value for $w$. Thus, knowing one side and the diagonal also requires 2 measurements ($l, d$).

If only one measurement is given (e.g., only the length of one side), the other dimension can vary, resulting in different rectangles.

Therefore, the minimum number of measurements required to construct a unique rectangle is 2.

The correct answer is (c) 2.

Solution:

In a quadrilateral, opposite sides are pairs of sides that do not share a common vertex.

For quadrilateral HOPE, the vertices are H, O, P, and E in sequence (or any order around the perimeter).

The pairs of opposite sides are formed by sides that do not share a common endpoint.

The sides are HO, OP, PE, and EH.

The pairs of opposite sides are (HO and PE) and (OP and EH).

The pairs of opposite sides are (HO, PE) and (OP, EH).

Solution:

In a quadrilateral, adjacent angles are two angles that share a common vertex and a common side.

For quadrilateral ROPE, the vertices are R, O, P, and E.

The angles are $\angle$R, $\angle$O, $\angle$P, and $\angle$E.

Adjacent angles are consecutive angles around the perimeter of the quadrilateral.

The pairs of adjacent angles are:

($\angle$R, $\angle$O) - sharing vertex O and side RO

($\angle$O, $\angle$P) - sharing vertex P and side OP

($\angle$P, $\angle$E) - sharing vertex E and side PE

($\angle$E, $\angle$R) - sharing vertex R and side ER

The pairs of adjacent angles are ($\angle$R, $\angle$O), ($\angle$O, $\angle$P), ($\angle$P, $\angle$E), and ($\angle$E, $\angle$R).

Solution:

In a quadrilateral, opposite angles are pairs of angles that are not adjacent (i.e., they do not share a common side).

For quadrilateral WXYZ, the vertices are W, X, Y, and Z.

The angles are $\angle$W, $\angle$X, $\angle$Y, and $\angle$Z.

The pairs of opposite angles are formed by angles that are across the quadrilateral from each other.

The pairs of opposite angles are ($\angle$W and $\angle$Y) and ($\angle$X and $\angle$Z).

The pairs of opposite angles are ($\angle$W, $\angle$Y) and ($\angle$X, $\angle$Z).

Solution:

A diagonal of a quadrilateral is a line segment connecting two non-adjacent vertices.

In quadrilateral DEFG, the vertices are D, E, F, and G.

The pairs of non-adjacent vertices are (D, F) and (E, G).

Therefore, the diagonals of the quadrilateral DEFG are the line segments connecting these pairs of vertices.

The diagonals are DF and EG.

The diagonals of the quadrilateral DEFG are DF and EG.

Solution:

A quadrilateral is a polygon with four sides and four interior angles.

The sum of the interior angles of any polygon with $n$ sides is given by the formula $(n-2) \times 180^\circ$.

For a quadrilateral, $n=4$.

Sum of interior angles = $(4-2) \times 180^\circ = 2 \times 180^\circ = 360^\circ$.

Therefore, the sum of all interior angles of a quadrilateral is 360°.

The sum of all interior angles of a quadrilateral is 360°.

Solution:

A regular pentagon is a polygon with 5 equal sides and 5 equal interior angles.

The sum of the exterior angles of any convex polygon is $360^\circ$.

For a regular polygon with $n$ sides, each exterior angle has the same measure.

The measure of each exterior angle of a regular $n$-sided polygon is given by the formula $\frac{360^\circ}{n}$.

For a regular pentagon, $n=5$.

Measure of each exterior angle = $\frac{360^\circ}{5}$

Calculation:

$360 \div 5 = 72$

So, the measure of each exterior angle of a regular pentagon is $72^\circ$.

The measure of each exterior angle of a regular pentagon is $\mathbf{72^\circ}$.

Solution:

A hexagon is a polygon with 6 sides.

The sum of the interior angles of a polygon with $n$ sides is given by the formula $(n-2) \times 180^\circ$.

For a hexagon, $n=6$.

Sum of the angles = $(6-2) \times 180^\circ$

Sum of the angles = $4 \times 180^\circ$

Calculation:

$4 \times 180 = 720$

So, the sum of the angles of a hexagon is $720^\circ$.

Sum of the angles of a hexagon is $\mathbf{720^\circ}$.

Solution:

A regular polygon with 18 sides is a polygon where all sides are equal in length and all interior angles are equal in measure.

The sum of the exterior angles of any convex polygon is $360^\circ$.

For a regular polygon with $n$ sides, each exterior angle has the same measure.

The measure of each exterior angle of a regular $n$-sided polygon is given by the formula:

$\text{Measure of each exterior angle} = \frac{360^\circ}{n}$

For a regular polygon of 18 sides, $n=18$.

Measure of each exterior angle = $\frac{360^\circ}{18}$

Calculation:

$\frac{360}{18} = 20$

So, the measure of each exterior angle of a regular polygon of 18 sides is $20^\circ$.

The measure of each exterior angle of a regular polygon of 18 sides is $\mathbf{20^\circ}$.

Solution:

For a regular polygon with $n$ sides, the measure of each exterior angle is given by the formula:

Measure of each exterior angle $= \frac{360^\circ}{n}$

We are given that the measure of each exterior angle is $36^\circ$.

So, we can set up the equation:

$36^\circ = \frac{360^\circ}{n}$

To find the number of sides, $n$, we can rearrange the equation:

$n = \frac{360^\circ}{36^\circ}$

Calculating the value:

$n = \frac{360}{36}$

$n = 10$

The number of sides of the regular polygon is 10.

The number of sides of a regular polygon, where each exterior angle has a measure of 36°, is 10.

Solution:

The description "a closed curve entirely made up of line segments" is the definition of a polygon.

A closed curve is one that starts and ends at the same point.

A line segment is a part of a line that has two endpoints.

When line segments are joined end-to-end to form a closed figure, it is called a polygon.

The another name for this shape is polygon.

Solution:

Let's consider different types of quadrilaterals and their properties regarding opposite angles:

1. Parallelogram: Both pairs of opposite angles are equal.

2. Rectangle: Both pairs of opposite angles are equal (all $90^\circ$).

3. Rhombus: Both pairs of opposite angles are equal.

4. Square: Both pairs of opposite angles are equal (all $90^\circ$).

5. Trapezium (Trapezoid): Opposite angles are generally not equal. An isosceles trapezium has one pair of opposite angles equal only if they are $90^\circ$, making it a rectangle.

6. Kite: A quadrilateral with two distinct pairs of equal adjacent sides. A kite has exactly one pair of opposite angles that are equal (the angles between the unequal sides).

The question asks for a quadrilateral that is "not a parallelogram" but has "exactly two opposite angles of equal measure".

A kite fits this description. In a kite, the angles between the unequal sides are equal, while the other pair of opposite angles (at the vertices where the equal sides meet) are generally unequal, unless the kite is also a rhombus or a square (which are parallelograms). Since the condition is "not a parallelogram", a general kite satisfies the criteria of having exactly two opposite angles equal.

A quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure is a kite.

Solution:

A regular pentagon is a polygon with 5 equal sides and 5 equal interior angles.

The sum of the interior angles of a polygon with $n$ sides is given by the formula $(n-2) \times 180^\circ$.

For a regular pentagon, $n=5$.

Sum of the interior angles = $(5-2) \times 180^\circ$

Sum of the interior angles = $3 \times 180^\circ$

Sum of the interior angles = $540^\circ$

Since it is a regular pentagon, each interior angle has the same measure.

Measure of each interior angle = $\frac{\text{Sum of interior angles}}{\text{Number of sides}}$

Measure of each interior angle = $\frac{540^\circ}{5}$

Calculation:

$540 \div 5 = 108$

So, the measure of each interior angle of a regular pentagon is $108^\circ$.

Alternatively, we can find the measure of the exterior angle first.

The measure of each exterior angle of a regular $n$-sided polygon is $\frac{360^\circ}{n}$.

For a regular pentagon, $n=5$.

Measure of each exterior angle = $\frac{360^\circ}{5} = 72^\circ$.

The interior angle and the exterior angle at each vertex form a linear pair, so their sum is $180^\circ$.

Measure of each interior angle = $180^\circ$ - Measure of each exterior angle

Measure of each interior angle = $180^\circ - 72^\circ = 108^\circ$.

The measure of each angle of a regular pentagon is $\mathbf{108^\circ}$.

Solution:

A polygon is a closed figure made up of line segments.

A polygon with three sides is called a triangle.

A regular polygon is a polygon in which all sides are equal in length and all interior angles are equal in measure.

Therefore, a three-sided regular polygon is a triangle with all sides equal and all angles equal.

This type of triangle is called an equilateral triangle.

The name of three-sided regular polygon is equilateral triangle.

Solution:

A hexagon is a polygon with 6 sides.

The number of diagonals in a polygon with $n$ sides can be calculated using the formula:

Number of diagonals = $\frac{n(n-3)}{2}$

For a hexagon, the number of sides $n = 6$.

Substituting $n=6$ into the formula:

Number of diagonals = $\frac{6(6-3)}{2}$

Number of diagonals = $\frac{6 \times 3}{2}$

Number of diagonals = $\frac{18}{2}$

Number of diagonals = $9$

So, there are 9 diagonals in a hexagon.

The number of diagonals in a hexagon is 9.

Solution:

The definition of a polygon is a simple closed curve formed by the union of a finite number of line segments, called sides, where each segment meets exactly two others at their endpoints, called vertices.

The question provides the beginning of this definition: "A polygon is a simple closed curve made up of only...".

Based on the definition, a polygon is made up of line segments.

A polygon is a simple closed curve made up of only line segments.

Solution:

A regular polygon is defined as a polygon that is both equilateral (all sides are equal) and equiangular (all angles are equal).

The provided sentence starts the definition:

"A regular polygon is a polygon whose all sides are equal and all __________ are equal."

To complete the definition of a regular polygon, the missing property is that all its angles are equal.

A regular polygon is a polygon whose all sides are equal and all angles are equal.

Solution:

The sum of the interior angles of a polygon with $n$ sides is given by the formula:

Sum of interior angles $= (n-2) \times 180^\circ$

A right angle measures $90^\circ$.

To express the sum of interior angles in terms of right angles, we need to divide the sum by the measure of a right angle ($90^\circ$).

Sum of interior angles in right angles $= \frac{(n-2) \times 180^\circ}{90^\circ}$

Simplify the expression:

Sum of interior angles in right angles $= (n-2) \times \frac{180}{90}$

Sum of interior angles in right angles $= (n-2) \times 2$

Sum of interior angles in right angles $= 2(n-2)$

The sum of interior angles of a polygon of n sides is $\mathbf{2(n-2)}$ right angles.

Solution:

For any convex polygon, the sum of the measures of its exterior angles, one at each vertex, is always constant, regardless of the number of sides.

This constant sum is equal to $360^\circ$.

The sum of all exterior angles of a polygon is $\mathbf{360^\circ}$.

Solution:

A quadrilateral is a polygon with four sides.

A regular polygon is a polygon that is both equilateral (all sides equal) and equiangular (all angles equal).

Therefore, a regular quadrilateral is a four-sided polygon with all four sides equal in length and all four interior angles equal in measure.

Let the side length be $s$. The quadrilateral has four sides of length $s$.

Let the angle measure be $\alpha$. The quadrilateral has four interior angles of measure $\alpha$.

The sum of the interior angles of a quadrilateral is $360^\circ$ (from Question 57 or the formula $(4-2) \times 180^\circ$).

Since all four angles are equal, we have:

$4\alpha = 360^\circ$

$\alpha = \frac{360^\circ}{4} = 90^\circ$

So, a regular quadrilateral must have four equal sides and four $90^\circ$ angles.

A quadrilateral with four equal sides is a rhombus.

A quadrilateral with four $90^\circ$ angles is a rectangle.

A quadrilateral that is both a rhombus and a rectangle is a square.

A square has four equal sides and four right angles ($90^\circ$).

Therefore, a square is a regular quadrilateral.

Square is a regular quadrilateral.

Solution:

A quadrilateral is a four-sided polygon.

The question describes a quadrilateral where at least one pair of opposite sides is parallel.

This is the definition of a trapezium (also known as a trapezoid in some regions).

In a trapezium, one or both pairs of opposite sides can be parallel.

If both pairs of opposite sides are parallel, the quadrilateral is a parallelogram, which is a special type of trapezium.

If only one pair of opposite sides is parallel, it is a non-parallelogram trapezium.

The most general term for a quadrilateral with at least one pair of parallel sides is a trapezium.

A quadrilateral in which a pair of opposite sides is parallel is a trapezium.

Solution:

A quadrilateral is a four-sided polygon.

The question specifies a quadrilateral where all four sides are equal in length.

Let the quadrilateral be ABCD, and let the side lengths be AB, BC, CD, and DA.

The condition is AB = BC = CD = DA.

A quadrilateral with all four sides equal is called a rhombus.

A square is a special case of a rhombus where all angles are also equal ($90^\circ$). However, if only the side equality is given, the quadrilateral is generally a rhombus.

If all sides of a quadrilateral are equal, it is a rhombus.

Solution:

A rhombus is a quadrilateral with all four sides equal in length.

One of the key properties of the diagonals of a rhombus is that they are perpendicular bisectors of each other.

When two line segments are perpendicular, they intersect at a $90^\circ$ angle.

Therefore, the diagonals of a rhombus intersect at right angles.

In a rhombus diagonals intersect at right angles.

Solution:

To determine a geometric figure uniquely, we need enough independent measurements to fix its size and shape.

A quadrilateral has 8 components: 4 sides and 4 interior angles. However, these components are not all independent.

For example, the sum of the interior angles is always $360^\circ$.

Consider constructing a quadrilateral. We can start by drawing a side, then an angle, then another side, and so on.

If we are given all four side lengths, the shape is not unique (e.g., a rhombus with given side length can be squeezed or stretched). We need additional information to fix the angles.

Similarly, if we are given all four angle measures (which sum to $360^\circ$), the shape is not unique; polygons with the same angles can be scaled up or down (e.g., different sized rectangles).

To uniquely determine a quadrilateral, we generally need 5 independent measurements.

Some common combinations of 5 measurements that can uniquely determine a quadrilateral are:

1. Four sides and one diagonal.

2. Three sides and two included angles.

3. Two adjacent sides and three angles.

These sets of measurements provide enough information to fix the relative positions of the vertices and thus the shape and size of the quadrilateral.

Five measurements can determine a quadrilateral uniquely.

Solution:

As discussed in the previous question (Q75), a quadrilateral generally requires 5 independent measurements for unique construction.

The question states that we are given three sides of the quadrilateral.

Let the number of sides given be $s = 3$.

Let the number of angles needed to complete the set of measurements for unique construction be $a$.

The total number of independent measurements required is 5.

So, $s + a = 5$.

$3 + a = 5$

$a = 5 - 3 = 2$.

Therefore, two angles are needed in addition to the three sides to uniquely construct a quadrilateral. A common scenario for this is when three consecutive sides and the two included angles are given (SASAS construction). The specific positions of the sides and angles are crucial for unique construction, but the number of angles needed to reach the required 5 measurements, when 3 sides are given, is 2.

A quadrilateral can be constructed uniquely if its three sides and two angles are given.

Solution:

A parallelogram is a quadrilateral with opposite sides parallel and equal in length.

A rhombus is defined as a quadrilateral with all four sides equal in length.

Alternatively, a rhombus can be defined as a parallelogram with adjacent sides equal.

If a parallelogram has adjacent sides equal, say side AB and BC are equal in parallelogram ABCD (where AB is parallel to CD and BC is parallel to AD).

Since it is a parallelogram, AB = CD and BC = AD.

If AB = BC, then AB = BC = CD = AD.

So, all four sides become equal.

The statement "A rhombus is a parallelogram in which __________ sides are equal" is describing the specific property of a parallelogram that makes it a rhombus.

The most fundamental side property of a rhombus is that all its sides are equal.

A rhombus is a parallelogram in which all sides are equal.

Solution:

A convex quadrilateral is a quadrilateral where all interior angles are less than $180^\circ$, and all diagonals lie entirely within the quadrilateral.

A concave quadrilateral is a quadrilateral that is not convex.

The defining characteristic of a concave polygon (which includes concave quadrilaterals) is that at least one of its interior angles has a measure greater than $180^\circ$. Such an angle is often called a reflex angle.

The statement says "The measure of __________ angle of concave quadrilateral is more than 180°." This refers to the specific angle within the concave quadrilateral that has this property.

This angle is one of the interior angles of the quadrilateral.

The measure of an interior angle of concave quadrilateral is more than 180°.

Solution:

In a polygon, vertices are the points where the sides meet.

Adjacent vertices are two vertices that are connected by a side of the polygon.

Non-adjacent vertices are two vertices that are not connected by a side of the polygon.

A diagonal is defined as a line segment connecting two non-adjacent vertices of a polygon.

For a quadrilateral with vertices A, B, C, D in order:

• A and B are adjacent.
• B and C are adjacent.
• C and D are adjacent.
• D and A are adjacent.

The non-adjacent pairs of vertices are (A, C) and (B, D).

The line segments joining these pairs, AC and BD, are the diagonals.

The term "opposite" vertices is also used to describe non-adjacent vertices in a quadrilateral.

A diagonal of a quadrilateral is a line segment that joins two non-adjacent vertices of the quadrilateral.

Alternatively,

A diagonal of a quadrilateral is a line segment that joins two opposite vertices of the quadrilateral.

Solution:

For a regular polygon with $n$ sides, the measure of each exterior angle is given by the formula:

Measure of each exterior angle $= \frac{360^\circ}{n}$

We are given that the measure of an exterior angle is $72^\circ$.

So, we can set up the equation:

$72^\circ = \frac{360^\circ}{n}$

To find the number of sides, $n$, we rearrange the equation:

$n = \frac{360^\circ}{72^\circ}$

Calculating the value:

$n = \frac{360}{72}$

$n = 5$

The number of sides of the regular polygon is 5.

The number of sides in a regular polygon having measure of an exterior angle as 72° is 5.

Solution:

Let the quadrilateral be ABCD, and let its diagonals be AC and BD. The diagonals intersect at a point, say O.

If the diagonals bisect each other, it means that the point of intersection O is the midpoint of both diagonals.

So, AO = OC and BO = OD.

Consider the triangles formed by the intersection of the diagonals, for example, $\triangle$AOB and $\triangle$COD.

In $\triangle$AOB and $\triangle$COD:

AO = OC

BO = OD

$\angle$AOB = $\angle$COD (Vertically opposite angles)

By the Side-Angle-Side (SAS) congruence criterion, $\triangle$AOB $\cong$ $\triangle$COD.

From the congruence, corresponding parts are equal (CPCT):

AB = CD

$\angle$OAB = $\angle$OCD

Since $\angle$OAB and $\angle$OCD are alternate interior angles formed by the transversal AC intersecting lines AB and CD, and they are equal, this implies that AB is parallel to CD.

Similarly, considering $\triangle$AOD and $\triangle$COB:

AO = OC

OD = OB

$\angle$AOD = $\angle$COB (Vertically opposite angles)

By the SAS congruence criterion, $\triangle$AOD $\cong$ $\triangle$COB.

From the congruence, corresponding parts are equal (CPCT):

AD = CB

$\angle$OAD = $\angle$OCB

Since $\angle$OAD and $\angle$OCB are alternate interior angles formed by the transversal AC intersecting lines AD and BC, and they are equal, this implies that AD is parallel to BC.

A quadrilateral with both pairs of opposite sides parallel is defined as a parallelogram.

Therefore, if the diagonals of a quadrilateral bisect each other, it is a parallelogram.

If the diagonals of a quadrilateral bisect each other, it is a parallelogram.

Given:

Length of one adjacent side of the parallelogram ($a$) = 5 cm

Length of the other adjacent side of the parallelogram ($b$) = 9 cm

To Find:

The perimeter of the parallelogram.

Solution:

In a parallelogram, opposite sides are equal in length.

If the adjacent sides are 5 cm and 9 cm, then the four sides of the parallelogram have lengths 5 cm, 9 cm, 5 cm, and 9 cm.

The perimeter of a polygon is the sum of the lengths of all its sides.

Perimeter of parallelogram = Sum of the lengths of its four sides

Perimeter = 5 cm + 9 cm + 5 cm + 9 cm

Perimeter = (5 + 9 + 5 + 9) cm

Perimeter = 28 cm

Alternatively, the perimeter of a parallelogram with adjacent sides of lengths $a$ and $b$ is given by the formula:

$\text{Perimeter} = 2(a+b)$

Substituting the given values $a = 5$ cm and $b = 9$ cm:

$\text{Perimeter} = 2(5 + 9)$ cm

$\text{Perimeter} = 2(14)$ cm

$\text{Perimeter} = 28$ cm

The perimeter of the parallelogram is 28 cm.

Solution:

A nonagon is a polygon. The names of polygons are based on the number of sides they have.

The prefix "nona-" indicates nine.

Therefore, a nonagon is a polygon with 9 sides.

A nonagon has nine sides.

Solution:

A rectangle is a special type of parallelogram where all four interior angles are right angles ($90^\circ$).

Let the rectangle be ABCD, with vertices A, B, C, and D in order. The diagonals are AC and BD.

Consider the triangles $\triangle$ABC and $\triangle$DCB.

AB = DC (Opposite sides of a parallelogram are equal)

BC = CB (Common side)

$\angle$ABC = $\angle$DCB = $90^\circ$ (Angles of a rectangle)

By the Side-Angle-Side (SAS) congruence criterion, $\triangle$ABC $\cong$ $\triangle$DCB.

From the congruence, corresponding parts are equal (CPCT):

AC = DB

This shows that the diagonals AC and BD are equal in length.

Also, since a rectangle is a parallelogram, its diagonals bisect each other.

So, the diagonals of a rectangle are equal in length and bisect each other.

The most specific property describing the equality of diagonals in a rectangle is that they are equal.

Diagonals of a rectangle are equal.

Alternatively,

Diagonals of a rectangle are equal and bisect each other.

Solution:

Polygons are named based on the number of sides they have.

A polygon with 3 sides is a triangle.

A polygon with 4 sides is a quadrilateral.

A polygon with 5 sides is a pentagon.

A polygon with 6 sides is a hexagon.

A polygon with 7 sides is a heptagon.

A polygon with 8 sides is an octagon.

A polygon with 9 sides is a nonagon.

A polygon with 10 sides is called a decagon.

A polygon having 10 sides is known as a decagon.

Solution:

A rectangle is a quadrilateral with four right angles. In a rectangle, opposite sides are equal in length.

Let the rectangle have adjacent sides of lengths $l$ and $w$. The perimeter sides would typically be $l, w, l, w$.

The additional condition given is that the adjacent sides are equal. So, $l = w$.

If $l = w$, then all four sides of the quadrilateral are equal in length ($l=w=l=w$).

So, the quadrilateral has:

1. Four right angles (because it is a rectangle).

2. Four equal sides (because adjacent sides are equal, and opposite sides are equal).

A quadrilateral with four equal sides is a rhombus.

A quadrilateral with four right angles is a rectangle.

A quadrilateral that is both a rhombus and a rectangle is a square.

Therefore, a rectangle whose adjacent sides are equal is a square.

A rectangle whose adjacent sides are equal becomes a square.

Given:

Length of one diagonal of a rectangle = 6 cm

To Find:

Length of the other diagonal.

Solution:

One of the properties of a rectangle is that its diagonals are equal in length.

Let $d_1$ and $d_2$ be the lengths of the two diagonals of the rectangle.

According to the property of a rectangle:

$d_1 = d_2$

We are given that the length of one diagonal is 6 cm.

Let $d_1 = 6$ cm.

Then, the length of the other diagonal is $d_2 = d_1 = 6$ cm.

If one diagonal of a rectangle is 6 cm long, length of the other diagonal is 6 cm.

Solution:

In a parallelogram, adjacent angles are consecutive angles that share a common side.

Consider a parallelogram ABCD. The pairs of adjacent angles are ($\angle$A, $\angle$B), ($\angle$B, $\angle$C), ($\angle$C, $\angle$D), and ($\angle$D, $\angle$A).

Since opposite sides of a parallelogram are parallel (e.g., AD || BC and AB || DC), we can consider a transversal line intersecting the parallel lines.

For example, consider parallel lines AD and BC intersected by transversal AB.

The angles $\angle$A and $\angle$B are consecutive interior angles on the same side of the transversal.

Therefore, their sum is $180^\circ$.

$\angle$A + $\angle$B = $180^\circ$

Similarly, for other pairs of adjacent angles:

$\angle$B + $\angle$C = $180^\circ$ (AD || BC, transversal BC)

$\angle$C + $\angle$D = $180^\circ$ (AB || DC, transversal CD)

$\angle$D + $\angle$A = $180^\circ$ (AB || DC, transversal DA)

Angles whose sum is $180^\circ$ are called supplementary angles.

Adjacent angles of a parallelogram are supplementary.

Solution:

Let the quadrilateral be ABCD, and let the diagonals AC and BD intersect at point O.

The statement "only one diagonal bisects the other" means that one diagonal cuts the other into two equal parts at their intersection point, but the reverse is not true for the first diagonal.

Let's examine the properties of diagonals in common quadrilaterals:

• In a parallelogram, the diagonals bisect each other. This means both diagonals are bisected. This does not fit the condition "only one diagonal... bisects the other".
• In a rectangle, the diagonals are equal and bisect each other. (Doesn't fit).
• In a rhombus, the diagonals are perpendicular bisectors of each other. (Doesn't fit).
• In a square, the diagonals are equal perpendicular bisectors of each other. (Doesn't fit).
• In a general trapezium, the diagonals do not necessarily bisect each other.
• In a kite, one of the diagonals is the perpendicular bisector of the other diagonal. This means that one diagonal is cut into two equal parts by the other diagonal, and they intersect at a right angle. The diagonal that is bisected is the one connecting the vertices where the two pairs of equal adjacent sides meet. The other diagonal (the axis of symmetry) bisects the angles, is the perpendicular bisector of the first diagonal, but is generally not bisected itself unless the kite is also a rhombus (in which case it's a parallelogram).

So, in a kite that is not a rhombus or a square, exactly one diagonal is bisected by the other.

Let's assume the kite has vertices A, B, C, D such that AB = AD and CB = CD. The diagonal AC is the axis of symmetry. The diagonals AC and BD intersect at O. In this case, AC is the perpendicular bisector of BD, meaning BO = OD and AC $\perp$ BD. However, AO is generally not equal to OC (unless A, B, C, D form a rhombus). Thus, diagonal AC bisects diagonal BD, but diagonal BD does not bisect diagonal AC.

This property exactly matches the condition "only one diagonal of a quadrilateral bisects the other".

If only one diagonal of a quadrilateral bisects the other, then the quadrilateral is known as a kite.

Given:

Trapezium ABCD with AB || CD.

Measure of angle A, $\angle$A = $100^\circ$.

To Find:

Measure of angle D, $\angle$D.

Solution:

In a trapezium, if a pair of opposite sides is parallel, then the consecutive interior angles on the same side of the transversal are supplementary (their sum is $180^\circ$).

Given that AB || CD, the transversal AD intersects these parallel lines.

The angles $\angle$A and $\angle$D are consecutive interior angles on the same side of the transversal AD.

Therefore, the sum of $\angle$A and $\angle$D is $180^\circ$.

Substitute the given value of $\angle$A = $100^\circ$ into the equation:

Subtract $100^\circ$ from both sides to solve for $\angle$D:

In trapezium ABCD with AB || CD, if ∠A = 100°, then ∠D = $\mathbf{80^\circ}$.

Solution:

Let the polygon have $n$ sides.

The sum of the measures of the exterior angles of any convex polygon is always $360^\circ$.

The sum of the measures of the interior angles of a polygon with $n$ sides is given by the formula $(n-2) \times 180^\circ$.

According to the problem statement, the sum of all exterior angles is equal to the sum of interior angles.

So, we have the equation:

Sum of exterior angles = Sum of interior angles

$360^\circ = (n-2) \times 180^\circ$

To find the number of sides $n$, we can solve this equation.

Divide both sides by $180^\circ$:

$\frac{360^\circ}{180^\circ} = n-2$

$2 = n-2$

Add 2 to both sides of the equation:

$n = 2 + 2$

$n = 4$

The polygon has 4 sides.

A polygon with 4 sides is called a quadrilateral.

The polygon in which sum of all exterior angles is equal to the sum of interior angles is called a quadrilateral.

False

In a trapezium, only one pair of opposite sides is parallel. The angles are generally not equal. For example, consider a right trapezium; it has two angles of $90^\circ$, but the other two angles can be different.

True

A square is a quadrilateral with four right angles and four equal sides. A rectangle is defined as a quadrilateral with four right angles. Since a square satisfies the definition of a rectangle, all squares are rectangles.

False

A kite is a quadrilateral with two distinct pairs of adjacent sides of equal length. A square is a quadrilateral with four equal sides and four right angles. While a square has two pairs of equal adjacent sides (all sides are equal), it also has the condition of having all right angles. A kite does not necessarily have all equal sides or all right angles. Therefore, not all kites are squares.

True

A rectangle is a quadrilateral with four right angles. In a quadrilateral with four right angles, opposite sides are parallel and equal in length. A parallelogram is defined as a quadrilateral with opposite sides parallel. Since a rectangle has opposite sides parallel, all rectangles are parallelograms.

False

A rhombus is a quadrilateral with all four sides equal in length. A square is a rhombus with all four angles equal to $90^\circ$. Not all rhombuses have right angles; therefore, not all rhombuses are squares. A rhombus that is not a square has unequal adjacent angles.

False

The sum of all the interior angles of a convex quadrilateral is $360^\circ$. This can be seen by dividing the quadrilateral into two triangles using a diagonal. The sum of angles in each triangle is $180^\circ$, so the total sum is $180^\circ + 180^\circ = 360^\circ$.

True

A quadrilateral is a polygon with four vertices. A diagonal is a line segment connecting two non-adjacent vertices. In a quadrilateral, there are two pairs of non-adjacent vertices, so there are exactly two diagonals.

True

The sum of the interior angles of a triangle is always $180^\circ$.

The sum of the exterior angles of any convex polygon (including a triangle), taking one at each vertex, is always $360^\circ$.

We need to check if the sum of exterior angles ($360^\circ$) is double the sum of interior angles ($180^\circ$).

$2 \times (\text{Sum of interior angles}) = 2 \times 180^\circ = 360^\circ$.

Since the sum of exterior angles is $360^\circ$, the statement is true for a triangle.

False

A polygon is a closed plane figure formed by a finite sequence of straight line segments. The given image shows a figure with curved boundaries, which are not straight line segments. Therefore, the figure is not a polygon.

False

A convex quadrilateral is defined as a quadrilateral where for every side, all other vertices lie on the same side of the line containing that side. Equivalently, a quadrilateral is convex if all its interior angles are less than $180^\circ$. Another way to check is that both diagonals lie entirely within the quadrilateral.

A kite is a quadrilateral where two distinct pairs of adjacent sides are equal in length. In a typical kite (the convex one), the vertices are arranged such that the kite bulges outwards. The two diagonals of a standard kite intersect inside the figure.

Since all interior angles of a standard kite are less than $180^\circ$ and its diagonals lie within the figure, it satisfies the conditions of a convex quadrilateral. While there exists a concave shape with two pairs of equal adjacent sides (often called a dart or arrowhead), the term 'kite' usually refers to the convex form in geometry unless specified otherwise.

True

Let $n$ be the number of sides of a polygon.

The sum of the interior angles of a convex polygon with $n$ sides is given by the formula $(n-2) \times 180^\circ$.

The sum of the exterior angles of any convex polygon, taken one at each vertex, is always $360^\circ$.

We want to find when the sum of interior angles is equal to the sum of exterior angles:

Sum of Interior Angles = Sum of Exterior Angles

$(n-2) \times 180^\circ = 360^\circ$

Divide both sides by $180^\circ$:

$n-2 = \frac{360^\circ}{180^\circ}$

$n-2 = 2$

Add 2 to both sides:

$n = 2 + 2$

$n = 4$

The only polygon for which the sum of interior angles equals the sum of exterior angles is a polygon with 4 sides, which is a quadrilateral. Thus, the statement is true.

True

Let the polygon have $n$ sides.

The sum of the interior angles of a convex polygon with $n$ sides is given by the formula:

Sum of Interior Angles = $(n-2) \times 180^\circ$

The sum of the exterior angles of any convex polygon, taken one at each vertex, is always:

Sum of Exterior Angles = $360^\circ$

According to the given condition, the sum of interior angles is double the sum of exterior angles:

Sum of Interior Angles = $2 \times$ Sum of Exterior Angles

$(n-2) \times 180^\circ = 2 \times 360^\circ$

$(n-2) \times 180^\circ = 720^\circ$

To find the number of sides $n$, we solve the equation:

$n-2 = \frac{720^\circ}{180^\circ}$

$n-2 = 4$

Adding 2 to both sides:

$n = 4 + 2$

$n = 6$

A polygon with 6 sides is called a hexagon. Since we found $n=6$, the polygon must be a hexagon. Thus, the statement is true.

False

A polygon is considered regular if and only if it is both equilateral (all sides are equal) and equiangular (all interior angles are equal).

Simply having all sides equal (equilateral) is not sufficient for a polygon to be regular. For example, a rhombus has all four sides equal, but its interior angles are not necessarily equal (unless it is also a square). A rhombus that is not a square is equilateral but not regular.

False

A regular polygon is a polygon that is both equilateral (all sides are equal) and equiangular (all interior angles are equal).

A quadrilateral is a polygon with four sides.

A rectangle is a quadrilateral with four right angles. This means a rectangle is always equiangular, as all its interior angles are $90^\circ$.

However, for a rectangle to be regular, it must also be equilateral, meaning all its sides must be equal. Rectangles only have opposite sides equal, not necessarily all four sides equal.

The only quadrilateral that is both equilateral and equiangular is a square. Therefore, a square is the only regular quadrilateral.

Since not all rectangles have equal sides, a rectangle is not always equilateral, and therefore, it is not a regular quadrilateral (unless it is a square).

False

While all rectangles have equal diagonals, having equal diagonals does not necessarily mean a quadrilateral is a rectangle.

A counterexample is an isosceles trapezoid. An isosceles trapezoid is a quadrilateral with at least one pair of parallel sides and equal non-parallel sides. The diagonals of an isosceles trapezoid are equal in length, but it is generally not a rectangle (as its angles are not all $90^\circ$).

True

Let the quadrilateral be ABCD, with angles $\angle A, \angle B, \angle C, \angle D$ at vertices A, B, C, and D respectively.

Given that opposite angles are equal:

The sum of the interior angles of a quadrilateral is $360^\circ$.

$\angle A + \angle B + \angle C + \angle D = 360^\circ$

Substitute (i) and (ii) into the sum of angles equation:

$\angle A + \angle B + \angle A + \angle B = 360^\circ$

$2\angle A + 2\angle B = 360^\circ$

$2(\angle A + \angle B) = 360^\circ$

$\angle A + \angle B = \frac{360^\circ}{2}$

Equation (iii) shows that the sum of adjacent angles $\angle A$ and $\angle B$ is $180^\circ$. When a transversal (like AB) intersects two lines (AD and BC) and the consecutive interior angles on the same side of the transversal are supplementary ($180^\circ$), the lines are parallel.

Thus, from $\angle A + \angle B = 180^\circ$, we can conclude that AD is parallel to BC ($AD || BC$).

Similarly, using (i), (ii), and (iii):

$\angle B + \angle C = \angle B + \angle A = 180^\circ$ (using (i))

This implies AB is parallel to DC ($AB || DC$).

Since both pairs of opposite sides are parallel ($AD || BC$ and $AB || DC$), the quadrilateral ABCD is a parallelogram.

Therefore, if opposite angles of a quadrilateral are equal, it must be a parallelogram.

False

Let the interior angles of the triangle be $x$, $2x$, and $3x$ according to the given ratio $1 : 2 : 3$.

The sum of the interior angles of a triangle is $180^\circ$.

$x + 2x + 3x = 180^\circ$

$6x = 180^\circ$

$x = \frac{180^\circ}{6}$

$x = 30^\circ$

So, the interior angles are:

Angle 1 = $x = 30^\circ$

Angle 2 = $2x = 2 \times 30^\circ = 60^\circ$}

Angle 3 = $3x = 3 \times 30^\circ = 90^\circ$}

The exterior angle at each vertex is supplementary to the interior angle at that vertex (their sum is $180^\circ$).

The exterior angles are:

Exterior Angle 1 = $180^\circ - 30^\circ = 150^\circ$}

Exterior Angle 2 = $180^\circ - 60^\circ = 120^\circ$}

Exterior Angle 3 = $180^\circ - 90^\circ = 90^\circ$}

The ratio of the exterior angles is $150 : 120 : 90$.

To simplify this ratio, we can divide each term by the greatest common divisor, which is 30.

Ratio = $\frac{150}{30} : \frac{120}{30} : \frac{90}{30}$}

Ratio = $5 : 4 : 3$

The ratio of the exterior angles is $5 : 4 : 3$, not $3 : 2 : 1$ as stated in the question.

Therefore, the statement is false.

True

A pentagon is a polygon with 5 sides.

A concave polygon (or non-convex polygon) is a polygon such that at least one of its interior angles is greater than $180^\circ$. Another way to identify a concave polygon is if at least one of its diagonals lies partially or entirely outside the polygon.

Assuming the image shows the standard representation of a concave pentagon from the source material (typically a 5-sided shape with one or more vertices pointing inwards, resulting in an interior angle greater than $180^\circ$), the figure is indeed a concave pentagon. The statement is true based on the visual properties of the figure shown.

False

A rhombus is a quadrilateral with all four sides equal in length.

Properties of the diagonals of a rhombus:

1. The diagonals bisect each other (since a rhombus is a type of parallelogram).

2. The diagonals are perpendicular to each other.

3. The diagonals bisect the angles at the vertices.

The statement claims that the diagonals are both equal and perpendicular. While the diagonals of a rhombus are always perpendicular, they are only equal in the specific case when the rhombus is also a square. A rhombus that is not a square has diagonals of unequal length.

Therefore, the statement is false because the diagonals of a rhombus are not always equal.

True

A rectangle is a quadrilateral with four right angles. Let the rectangle be ABCD, with vertices in order.

The diagonals are AC and BD.

Consider triangles ABC and DCB.

We know that:

By the SAS (Side-Angle-Side) congruence criterion, $\triangle ABC \cong \triangle DCB$.

Since the triangles are congruent, their corresponding parts are equal. The hypotenuses of these right triangles are the diagonals of the rectangle.

Thus, the diagonals of a rectangle are equal.

False

A rectangle is a parallelogram, and the diagonals of any parallelogram bisect each other.

Also, the diagonals of a rectangle are equal in length.

However, the diagonals of a rectangle bisect each other at right angles only if the rectangle is also a rhombus (i.e., it is a square). For a rectangle that is not a square, the diagonals intersect at an angle that is not $90^\circ$.

Since not all rectangles are squares, the statement that the diagonals of a rectangle bisect each other at right angles is false.

False

A kite is a quadrilateral with two distinct pairs of adjacent sides of equal length.

A parallelogram is a quadrilateral with both pairs of opposite sides parallel (or equal in length).

In a kite, the adjacent sides are equal, not necessarily the opposite sides. For a kite to be a parallelogram, it must have opposite sides equal or parallel. This happens only in the special case where all four sides are equal, making it a rhombus. A rhombus is a parallelogram.

However, not all kites are rhombuses. A typical kite has two pairs of equal adjacent sides, but the lengths of the sides in one pair are different from the lengths of the sides in the other pair. Such a kite does not have opposite sides of equal length or opposite sides parallel.

Therefore, every kite is not a parallelogram. Only a rhombus (which is a special type of kite where all four sides are equal) is also a parallelogram.

False

A trapezium (or trapezoid) is a quadrilateral with at least one pair of parallel sides.

A parallelogram is a quadrilateral with both pairs of opposite sides parallel.

For a trapezium to be a parallelogram, it must have two pairs of parallel sides. However, a typical trapezium has only one pair of parallel sides.

Therefore, not every trapezium satisfies the definition of a parallelogram. For example, a shape with exactly one pair of parallel sides is a trapezium but not a parallelogram.

(Note: Conversely, every parallelogram is a trapezium because it has at least one pair of parallel sides, in fact two.)

False

A parallelogram is a quadrilateral with both pairs of opposite sides parallel.

A rectangle is a parallelogram with four right angles.

While all rectangles are parallelograms (because opposite sides are parallel), not all parallelograms are rectangles. A parallelogram only becomes a rectangle if its interior angles are all $90^\circ$. A parallelogram that is not a rectangle has angles that are not all right angles (it has two acute angles and two obtuse angles).

For example, a rhombus that is not a square is a parallelogram but not a rectangle. Similarly, a parallelogram with unequal adjacent sides and non-right angles is a parallelogram but not a rectangle.

Therefore, every parallelogram is not a rectangle.

False

A trapezium (or trapezoid) is a quadrilateral with at least one pair of parallel sides.

A rectangle is a quadrilateral with four right angles.

For a trapezium to be a rectangle, it must have four right angles. However, most trapeziums do not have four right angles. A rectangle is a special type of trapezium (as it has two pairs of parallel sides, fulfilling the "at least one pair" condition), but not all trapeziums are rectangles.

For example, a trapezium with only one pair of parallel sides and non-right angles is a trapezium but not a rectangle.

Therefore, every trapezium is not a rectangle.

True

A rectangle is a quadrilateral with four right angles. In a rectangle, opposite sides are parallel.

A trapezium (or trapezoid) is a quadrilateral with at least one pair of parallel sides.

Since a rectangle has two pairs of parallel sides (opposite sides are parallel), it satisfies the condition of having "at least one pair of parallel sides". Therefore, every rectangle is a trapezium according to the definition of a trapezium as having at least one pair of parallel sides.

True

A square is a quadrilateral with four equal sides and four right angles.

A rhombus is a quadrilateral with all four sides equal in length.

Since a square has all four sides equal, it satisfies the definition of a rhombus. Therefore, every square is a rhombus.

(Note: The difference is that a rhombus does not necessarily have right angles, while a square does. A square is a special type of rhombus that is also a rectangle.)

True

A square is a quadrilateral with four equal sides and four right angles.

A parallelogram is a quadrilateral with both pairs of opposite sides parallel.

In a square, since all angles are $90^\circ$, consecutive interior angles sum up to $90^\circ + 90^\circ = 180^\circ$. This implies that opposite sides are parallel. Also, in a square, opposite sides are equal.

Since a square has opposite sides parallel (or equal), it satisfies the definition of a parallelogram. Therefore, every square is a parallelogram.

True

A square is a quadrilateral with four equal sides and four right angles.

As a property of having four right angles or being a parallelogram, a square has both pairs of opposite sides parallel.

A trapezium (or trapezoid) is a quadrilateral with at least one pair of parallel sides.

Since a square has two pairs of parallel sides, it certainly has at least one pair of parallel sides. Therefore, every square satisfies the definition of a trapezium.

True

A rhombus is a quadrilateral with all four sides equal in length. A property of a rhombus (as it is a type of parallelogram) is that its opposite sides are parallel. Thus, a rhombus has two pairs of parallel sides.

A trapezium (or trapezoid) is a quadrilateral with at least one pair of parallel sides.

Since a rhombus has two pairs of parallel sides, it satisfies the condition of having "at least one pair of parallel sides". Therefore, every rhombus is a trapezium.

False

To uniquely construct a quadrilateral, simply knowing the lengths of the four sides is not enough. Quadrilaterals are generally not rigid structures; they can be deformed while keeping the side lengths fixed.

For example, consider a rhombus. All four sides are equal. However, you can draw rhombuses with the same side lengths but different angles (and thus different shapes, areas, and diagonal lengths). A square and a non-square rhombus can have the same side length.

To uniquely define a quadrilateral, you need additional information, such as the measure of an angle or the length of a diagonal. The minimum information required to uniquely construct a quadrilateral is typically five independent pieces of information (e.g., 4 sides and 1 angle, or 3 sides and 2 angles, or 2 sides and 3 angles, or 4 sides and 1 diagonal, or 3 sides and 2 diagonals).

False

An obtuse angle is an angle whose measure is greater than $90^\circ$ but less than $180^\circ$.

Let the four interior angles of the quadrilateral be $\angle A, \angle B, \angle C, \angle D$.

The sum of the interior angles of a convex quadrilateral is $360^\circ$.

$\angle A + \angle B + \angle C + \angle D = 360^\circ$

If all four angles were obtuse, then each angle would be greater than $90^\circ$.

$\angle A > 90^\circ$

$\angle B > 90^\circ$

$\angle C > 90^\circ$

$\angle D > 90^\circ$

Summing these inequalities:

$\angle A + \angle B + \angle C + \angle D > 90^\circ + 90^\circ + 90^\circ + 90^\circ$

$\angle A + \angle B + \angle C + \angle D > 360^\circ$

This contradicts the fact that the sum of the interior angles of a quadrilateral is $360^\circ$. Therefore, it is impossible for a convex quadrilateral to have all four angles as obtuse.

(Note: A concave quadrilateral can have one reflex angle (greater than $180^\circ$), but even then, the sum of the four interior angles remains $360^\circ$. If one angle is reflex, the other three must be acute or right angles to keep the sum at $360^\circ$. So, a concave quadrilateral also cannot have four obtuse angles).

True

To construct a quadrilateral, knowing the lengths of the four sides and one diagonal provides sufficient information to uniquely determine the shape (up to congruence).

A diagonal divides the quadrilateral into two triangles. If we know the lengths of the four sides and one diagonal, say the diagonal is AC and the sides are AB, BC, CD, and DA, then we have the side lengths for two triangles: $\triangle ABC$ (sides AB, BC, AC) and $\triangle ADC$ (sides AD, DC, AC).

A triangle is uniquely determined by the lengths of its three sides (SSS congruence criterion). Therefore, $\triangle ABC$ can be uniquely constructed, and $\triangle ADC$ can also be uniquely constructed.

By joining these two triangles along the common diagonal (AC), the quadrilateral ABCD is uniquely formed.

Thus, a quadrilateral can be uniquely drawn if all four sides and one diagonal are known.

False

To draw a unique quadrilateral, knowing the measures of all four interior angles and the length of one side is not sufficient. The sum of the interior angles of any quadrilateral is always $360^\circ$. If you are given four angles, their sum must be $360^\circ$ for any such quadrilateral to exist.

However, knowing the angles only determines the shape (or aspect ratio) of the quadrilateral, not its actual size. For example, all rectangles have four $90^\circ$ angles. If you are given that a quadrilateral has four $90^\circ$ angles and one side length of 5 units, the adjacent side can be any length (e.g., 3 units, 7 units, etc.). This allows for the construction of infinitely many rectangles of different sizes, all with the same angles and one side length of 5.

Similarly, if you are given the angles of a parallelogram and the length of one side, the length of the adjacent side can vary, leading to different parallelograms with the same angles but different dimensions.

To uniquely determine a quadrilateral, you generally need five independent pieces of information (e.g., four sides and one diagonal, three sides and two angles, etc.). Knowing four angles gives 3 independent pieces of information (since the fourth is determined by the sum being $360^\circ$), and adding one side gives a total of 4 pieces of information, which is not enough for unique construction.

True

To uniquely determine a quadrilateral, you generally need five independent pieces of information.

If you are given the lengths of all four sides and the measure of one interior angle, this provides $4 + 1 = 5$ pieces of information.

Consider a quadrilateral ABCD with side lengths AB, BC, CD, DA, and the angle at vertex A ($\angle A$) is known.

You can construct triangle ABD using side AB, side AD, and the included angle $\angle A$. By the SAS congruence criterion, this triangle is uniquely determined. The length of the diagonal BD is thus uniquely determined.

Now consider triangle BCD. We know the lengths of sides BC and CD (given), and we have determined the length of side BD (the diagonal). By the SSS congruence criterion, triangle BCD is uniquely determined.

Since both triangles ABD and BCD are uniquely determined, the quadrilateral formed by joining them along the diagonal BD is also uniquely determined.

Therefore, a quadrilateral can be uniquely drawn if all four sides and one angle are known.

True

To uniquely determine a quadrilateral, you need five independent pieces of information.

If you are given the lengths of three sides and two diagonals, this provides $3 + 2 = 5$ pieces of information.

Let the quadrilateral be ABCD. Suppose the given sides are AB, BC, CD, and the given diagonals are AC and BD. This combination (3 sides + 2 diagonals) is not guaranteed to uniquely define a quadrilateral because the sides must be consecutive. Let's assume the given sides are three consecutive sides, say AB, BC, CD, and the two diagonals are AC and BD.

We know the lengths AB, BC, and diagonal AC. We can construct triangle ABC using SSS. This uniquely determines the positions of A, B, and C, and the angle $\angle ABC$.

Now, we know side CD. Vertex D must be at a distance equal to the length of CD from vertex C. So D lies on a circle centered at C with radius CD.

We also know the length of the diagonal BD. Vertex D must be at a distance equal to the length of BD from vertex B. So D also lies on a circle centered at B with radius BD.

The intersection points of these two circles will give the possible locations for vertex D. Typically, there will be at most two such intersection points. However, one of these points will be on one side of the line BC, and the other on the opposite side. Since a quadrilateral has its vertices in order, there is usually only one valid position for D on the correct side relative to A, B, and C.

Thus, knowing three consecutive sides and the two diagonals allows for the unique construction of the quadrilateral.

True

Let the quadrilateral be ABCD, and let the diagonals AC and BD intersect at point O. Given that the diagonals bisect each other, we have:

Consider triangles $\triangle AOB$ and $\triangle COD$.

We have:

By the SAS (Side-Angle-Side) congruence criterion, $\triangle AOB \cong \triangle COD$.

Since the triangles are congruent, their corresponding parts are equal:

The angles $\angle BAO$ (or $\angle BAC$) and $\angle DCO$ (or $\angle DCA$) are alternate interior angles formed by the transversal AC intersecting lines AB and DC. Since these alternate interior angles are equal, the lines AB and DC must be parallel ($AB || DC$).

Now consider triangles $\triangle AOD$ and $\triangle COB$.

We have:

By the SAS congruence criterion, $\triangle AOD \cong \triangle COB$.

Since the triangles are congruent, their corresponding parts are equal:

The angles $\angle DAO$ (or $\angle DAC$) and $\angle BCO$ (or $\angle BCA$) are alternate interior angles formed by the transversal AC intersecting lines AD and BC. Since these alternate interior angles are equal, the lines AD and BC must be parallel ($AD || BC$).

Since both pairs of opposite sides are parallel ($AB || DC$ and $AD || BC$), the quadrilateral ABCD is a parallelogram.

Therefore, if the diagonals of a quadrilateral bisect each other, it must be a parallelogram.

False

To uniquely construct a quadrilateral, we generally need five independent pieces of information (which are lengths or angles).

In this case, we are given three angles and the lengths of two sides. The sum of the interior angles of a quadrilateral is $360^\circ$, so knowing three angles automatically determines the fourth angle. Thus, knowing three angles is equivalent to having three independent pieces of information (the measures of the four angles, constrained by their sum).

Adding the lengths of two sides gives a total of $3 + 2 = 5$ pieces of information. While this is the required number, the combination of information is crucial for unique construction.

If the two given sides are adjacent and the three given angles are appropriately positioned relative to these sides (e.g., two angles adjacent to one side and one angle adjacent to the other, or two angles adjacent to both sides), the quadrilateral can likely be uniquely constructed.

However, the statement says "any two sides". Consider the case where the two given sides are opposite sides, and the three given angles are any three angles. For example, suppose we are given the lengths of sides AB and CD, and angles $\angle A, \angle B, \angle C$. Angle $\angle D$ is then also determined.

Consider a parallelogram. We know opposite sides are equal and opposite angles are equal. If we are given the lengths of two opposite sides (e.g., AB=CD=5) and the angles (e.g., $\angle A = 60^\circ, \angle B = 120^\circ, \angle C = 60^\circ, \angle D = 120^\circ$), we can construct infinitely many parallelograms with sides 5 and varying adjacent side lengths. The shape is determined by the angles, but the size is not fixed by just one pair of opposite sides.

Since there is at least one case (when the given sides are opposite) where the construction is not unique, the statement is false.

True

Let the lengths of the two diagonals of the parallelogram be $d_1$ and $d_2$, and let the angle between them at their intersection point be $\theta$.

We know that the diagonals of a parallelogram bisect each other. Let the diagonals be AC and BD, intersecting at point O.

Then $AO = OC = \frac{d_1}{2}$ and $BO = OD = \frac{d_2}{2}$.

We are given the angle of intersection, say $\angle AOB = \theta$. Consequently, the vertically opposite angle $\angle COD = \theta$. The adjacent angles are $\angle BOC = \angle DOA = 180^\circ - \theta$.

Consider triangle $\triangle AOB$. We know the lengths of two sides AO and BO (which are $\frac{d_1}{2}$ and $\frac{d_2}{2}$) and the measure of the included angle $\angle AOB = \theta$.

By the SAS (Side-Angle-Side) congruence criterion, a triangle is uniquely determined if two sides and the included angle are known.

Thus, $\triangle AOB$ is uniquely constructed. The length of the side AB is therefore uniquely determined.

Similarly, consider $\triangle BOC$. We know the lengths BO and CO ($\frac{d_2}{2}$ and $\frac{d_1}{2}$) and the included angle $\angle BOC = 180^\circ - \theta$. This triangle is also uniquely constructed, determining the length of side BC.

Since opposite sides of a parallelogram are equal, $CD = AB$ and $DA = BC$. The construction of $\triangle AOB$ and $\triangle BOC$ (sharing the vertex O and side OB/OA) uniquely fixes the positions of vertices A, B, C, and O relative to each other.

Vertex D is located such that OD = BO and CO = AO. The position of D relative to O is fixed by the segment BO and the angle $\theta$. Since $\triangle COD$ is congruent to $\triangle AOB$ and $\triangle DOA$ is congruent to $\triangle BOC$, the entire parallelogram ABCD is uniquely determined once the lengths of the half-diagonals and the angle between them are known.

Therefore, a parallelogram can be constructed uniquely if both diagonals and the angle between them is given.

True

A rhombus is a quadrilateral with all four sides equal.

Important properties of the diagonals of a rhombus:

1. They bisect each other.

2. They are perpendicular to each other.

Let the lengths of the two diagonals be $d_1$ and $d_2$. Let the diagonals intersect at point O. Since the diagonals bisect each other, the lengths of the segments from the center O to the vertices are $\frac{d_1}{2}$ and $\frac{d_2}{2}$.

Because the diagonals are perpendicular, they form four right-angled triangles at the center (e.g., $\triangle AOB$). In each of these right triangles, the two legs are the half-diagonals, and the hypotenuse is a side of the rhombus.

Consider one of these triangles, say $\triangle AOB$. We know the lengths of the two sides meeting at the right angle: $AO = \frac{d_1}{2}$ and $BO = \frac{d_2}{2}$. The angle between them is $\angle AOB = 90^\circ$.

By the SAS (Side-Angle-Side) congruence criterion, a triangle is uniquely determined if two sides and the included angle are known. Since $\frac{d_1}{2}$, $\frac{d_2}{2}$, and $90^\circ$ are fixed values, $\triangle AOB$ is uniquely constructed.

Once $\triangle AOB$ is uniquely constructed, the length of the side AB is determined (by the Pythagorean theorem, $AB = \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2}$). Since all sides of a rhombus are equal, this determines the length of all four sides.

Furthermore, the relative positions of vertices A and B with respect to the intersection O are fixed. By extending the segments AO and BO to C and D such that $OC=AO$ and $OD=BO$, and maintaining the $90^\circ$ intersection angle, the positions of C and D are also uniquely determined.

Thus, given the lengths of the two diagonals, a rhombus can be constructed uniquely.

Given:

Length of diagonal 1, $d_1 = 8$ cm.

Length of diagonal 2, $d_2 = 15$ cm.

To Find:

The length of the side of the rhombus.

Solution:

Let the rhombus be ABCD, and let the diagonals AC and BD intersect at point O.

We know that the diagonals of a rhombus bisect each other at right angles.

So, $AO = OC = \frac{d_1}{2}$ and $BO = OD = \frac{d_2}{2}$.

And $\angle AOB = 90^\circ$.

Using the given values:

$AO = \frac{8}{2} = 4$ cm

$BO = \frac{15}{2} = 7.5$ cm

Consider the right-angled triangle $\triangle AOB$. The sides AO and BO are the legs, and the side AB of the rhombus is the hypotenuse.

By the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides:

$AB^2 = AO^2 + BO^2$

Substitute the values of AO and BO:

$AB^2 = (4)^2 + (7.5)^2$

$AB^2 = 16 + (7.5 \times 7.5)$

$AB^2 = 16 + 56.25$

$AB^2 = 72.25$}

To find the side AB, take the square root of 72.25:

$AB = \sqrt{72.25}$

$AB = 8.5$ cm

Since all sides of a rhombus are equal, the side of the rhombus is 8.5 cm.

Given:

Two adjacent angles of a parallelogram are in the ratio $1:3$.

To Find:

The measures of all angles of the parallelogram.

Solution:

Let the two adjacent angles of the parallelogram be $x$ and $3x$, based on the given ratio $1:3$.

We know that adjacent angles of a parallelogram are supplementary, meaning their sum is $180^\circ$.

So,

$4x = 180^\circ$

Divide both sides by 4:

$x = \frac{180^\circ}{4}$

$x = 45^\circ$

The measures of the two adjacent angles are:

First angle = $x = 45^\circ$

Second angle = $3x = 3 \times 45^\circ = 135^\circ$

We also know that opposite angles of a parallelogram are equal.

So, the angle opposite to the first angle ($45^\circ$) is also $45^\circ$.

The angle opposite to the second angle ($135^\circ$) is also $135^\circ$.

The four angles of the parallelogram are $45^\circ$, $135^\circ$, $45^\circ$, and $135^\circ$.

Verification: The sum of the angles is $45^\circ + 135^\circ + 45^\circ + 135^\circ = 180^\circ + 180^\circ = 360^\circ$, which is correct for a quadrilateral.

The quadrilateral that is somewhat different from the others is the trapezium.

Justification:

Let's consider the definitions of the given quadrilaterals based on parallel sides:

1. A square is a parallelogram with four right angles and four equal sides. It has two pairs of parallel sides.

2. A rectangle is a parallelogram with four right angles. It has two pairs of parallel sides.

3. A rhombus is a parallelogram with four equal sides. It has two pairs of parallel sides.

4. A trapezium is a quadrilateral with at least one pair of parallel sides. In its most general form, it has exactly one pair of parallel sides.

Square, rectangle, and rhombus are all specific types of parallelograms. The defining characteristic of a parallelogram is having two pairs of parallel sides.

A trapezium, on the other hand, is defined by having *at least* one pair of parallel sides. A parallelogram fits this definition (it has two pairs, thus at least one). However, a trapezium can have exactly one pair of parallel sides, in which case it is not a parallelogram.

The square, rectangle, and rhombus form a group of quadrilaterals that are all parallelograms, sharing the property of having two pairs of parallel sides. The trapezium is different because it only requires one pair of parallel sides and, in its most common form, lacks the second pair of parallel sides that characterizes the other three.

Given:

Rectangle ABCD.

AB = 25 cm

BC = 15 cm

To Find:

The ratio in which the bisector of $\angle$C divides side AB.

Solution:

In rectangle ABCD, AB is parallel to DC and AD is parallel to BC. All interior angles are $90^\circ$.

We are given AB = 25 cm and BC = 15 cm.

Since it is a rectangle, AD = BC = 15 cm and DC = AB = 25 cm.

Let CE be the angle bisector of $\angle$C, where E is a point on side AB.

Since $\angle$BCD is an angle of a rectangle, $\angle$BCD = $90^\circ$.

The bisector CE divides $\angle$BCD into two equal angles:

Consider the triangle $\triangle$EBC.

$\angle EBC = 90^\circ$

From (i), $\angle BCE = 45^\circ$

The sum of angles in $\triangle$EBC is $180^\circ$.

$\angle BEC + \angle EBC + \angle BCE = 180^\circ$

$\angle BEC + 90^\circ + 45^\circ = 180^\circ$

$\angle BEC = 180^\circ - 90^\circ - 45^\circ$

In $\triangle$EBC, from (i) and (ii), we have $\angle BCE = \angle BEC = 45^\circ$.

Since the angles opposite to sides EB and BC are equal, the sides must be equal.

We are given BC = 15 cm.

So, EB = 15 cm.

The point E lies on the side AB. Therefore, AB = AE + EB.

We know AB = 25 cm and EB = 15 cm.

$25 = AE + 15$

$AE = 25 - 15$

$AE = 10$ cm

The bisector of $\angle$C divides AB into segments AE and EB.

The ratio AE : EB is $10 : 15$.

Simplifying the ratio by dividing both terms by their greatest common divisor (5):

$10 \div 5 = 2$

$15 \div 5 = 3$}

The ratio is $2 : 3$.

The bisector of $\angle$C divides AB in the ratio 2 : 3.

Given:

PQRS is a rectangle.

ST $\perp$ PR, where T is a point on PR.

ST divides $\angle$S in the ratio $2:3$, meaning $\angle PST : \angle RST = 2:3$.

To Find:

The measure of $\angle$TPQ.

Solution:

In rectangle PQRS, all interior angles are $90^\circ$. Thus, $\angle PSR = 90^\circ$.}

The perpendicular ST divides $\angle$S (which is $\angle$PSR) in the ratio $2:3$. The sum of the ratio parts is $2+3=5$.

So, $\angle PST = \frac{2}{5} \times \angle PSR = \frac{2}{5} \times 90^\circ$}

And $\angle RST = \frac{3}{5} \times \angle PSR = \frac{3}{5} \times 90^\circ$}

Since ST $\perp$ PR, $\angle STP = 90^\circ$.}

Consider the right-angled triangle $\triangle$STP. The sum of angles in a triangle is $180^\circ$.

$\angle STP + \angle PST + \angle SPT = 180^\circ$}

$126^\circ + \angle SPT = 180^\circ$

$\angle SPT = 180^\circ - 126^\circ$}

Note that $\angle SPT$ is the same as $\angle SPR$ (angle formed by diagonal PR with side PS).

In rectangle PQRS, $\angle QPS = 90^\circ$.}

The angle $\angle$QPS is the sum of $\angle$QPT (or $\angle$TPQ) and $\angle$SPT (or $\angle$SPR).

$\angle QPS = \angle TPQ + \angle SPR$}

Subtract $54^\circ$ from both sides:

$\angle TPQ = 90^\circ - 54^\circ$}

$\angle TPQ = 36^\circ$

The measure of $\angle$TPQ is $36^\circ$.

No, it is not a rectangle.

Justification:

One of the defining properties of a rectangle is that its diagonals are equal in length.

The given information states that the quadrilateral has one diagonal longer than the other. This means the lengths of the two diagonals are not equal.

Since the diagonals of the photo frame are unequal, it cannot be a rectangle because a rectangle must have equal diagonals.

Examples of quadrilaterals with unequal diagonals include kites, rhombuses (that are not squares), and general parallelograms (that are not rectangles or squares).

Given:

Adjacent angles of a parallelogram are $(2x – 4)^\circ$ and $(3x – 1)^\circ$.

To Find:

The measures of all angles of the parallelogram.

Solution:

We know that adjacent angles of a parallelogram are supplementary, meaning their sum is $180^\circ$.

So,

$(2x – 4)^\circ + (3x – 1)^\circ = 180^\circ$

Combine like terms:

$(2x + 3x) + (-4 - 1) = 180$

$5x - 5 = 180$

Add 5 to both sides:

$5x = 180 + 5$

$5x = 185$}

Divide both sides by 5:

$x = \frac{185}{5}$

$x = 37$

Now, substitute the value of $x$ into the expressions for the angles:

First angle = $(2x - 4)^\circ = (2 \times 37 - 4)^\circ = (74 - 4)^\circ = 70^\circ$

Second angle = $(3x - 1)^\circ = (3 \times 37 - 1)^\circ = (111 - 1)^\circ = 110^\circ$

The two adjacent angles are $70^\circ$ and $110^\circ$.

We know that opposite angles of a parallelogram are equal.

So, the angle opposite to $70^\circ$ is $70^\circ$.

The angle opposite to $110^\circ$ is $110^\circ$.

The four angles of the parallelogram are $70^\circ, 110^\circ, 70^\circ, 110^\circ$.

Verification: The sum of the angles is $70^\circ + 110^\circ + 70^\circ + 110^\circ = 180^\circ + 180^\circ = 360^\circ$, which is correct for a quadrilateral. Also, adjacent angles sum to $70^\circ + 110^\circ = 180^\circ$, confirming they are supplementary.

No, it cannot be a parallelogram.

Justification:

A fundamental property of a parallelogram is that its diagonals bisect each other. This means that the point of intersection of the two diagonals divides each diagonal into two equal segments. In other words, the ratio in which the intersection point divides each diagonal is $1:1$.

The problem states that the point of intersection of the diagonals divides *one* diagonal in the ratio $1:2$. This ratio ($1:2$) is not $1:1$.

Since the given condition contradicts the property that diagonals of a parallelogram bisect each other (divide in a $1:1$ ratio), the quadrilateral cannot be a parallelogram.

Given:

The ratio of the exterior angle to the interior angle of a regular polygon is $1:5$.

To Find:

The number of sides of the polygon.

Solution:

Let the measure of the exterior angle of the regular polygon be $x^\circ$ and the measure of the interior angle be $5x^\circ$, according to the ratio $1:5$.

We know that for any polygon, the exterior angle at a vertex and the interior angle at the same vertex are supplementary (their sum is $180^\circ$).

So,

Exterior Angle + Interior Angle = $180^\circ$

$x^\circ + 5x^\circ = 180^\circ$

$6x^\circ = 180^\circ$

Divide both sides by 6:

$x = \frac{180}{6}$

$x = 30$

So, the measure of the exterior angle is $30^\circ$, and the measure of the interior angle is $5 \times 30^\circ = 150^\circ$.

For a regular polygon with $n$ sides, the measure of each exterior angle is given by $\frac{360^\circ}{n}$.

We found the exterior angle to be $30^\circ$.

So,

To find $n$, rearrange the equation:

$n = \frac{360^\circ}{30^\circ}$

$n = \frac{36}{3}$

$n = 12$

The number of sides of the polygon is 12.

Alternatively, for a regular polygon with $n$ sides, the measure of each interior angle is given by $\frac{(n-2) \times 180^\circ}{n}$.

We found the interior angle to be $150^\circ$.

So,

Multiply both sides by $n$:

$(n-2) \times 180 = 150n$

Distribute 180:

$180n - 360 = 150n$

Subtract $150n$ from both sides:

$180n - 150n - 360 = 0$

$30n - 360 = 0$

Add 360 to both sides:

$30n = 360$

Divide both sides by 30:

$n = \frac{360}{30}$

$n = 12$

Both methods give the same result. The number of sides of the polygon is 12.

The shape formed by joining their end points is a rectangle.

Reason:

Let the two sticks represent the diagonals of the quadrilateral formed by joining their endpoints. Let the sticks be AC and BD, and let their intersection point be O.

Given that the sticks are each 5 cm in length, the diagonals of the quadrilateral have lengths AC = 5 cm and BD = 5 cm.

Thus, the diagonals are equal in length.

Given that the sticks bisect each other, the intersection point O is the midpoint of both diagonals. This means AO = OC and BO = OD.

A quadrilateral whose diagonals bisect each other is a parallelogram.

A parallelogram whose diagonals are equal in length is a rectangle.

Since the quadrilateral formed has diagonals that are both equal in length and bisect each other, it satisfies the properties of a rectangle.

The shape formed by joining their end points is a square.

Reason:

Let the two sticks represent the diagonals of the quadrilateral formed by joining their endpoints. Let the sticks be AC and BD, and let their intersection point be O.

Given that the sticks are each 7 cm in length, the diagonals of the quadrilateral have lengths AC = 7 cm and BD = 7 cm.

Thus, the diagonals are equal in length.

Given that the sticks bisect each other, the intersection point O is the midpoint of both diagonals. This means AO = OC and BO = OD.

A quadrilateral whose diagonals bisect each other is a parallelogram.

Given that the sticks bisect each other at right angles, the diagonals are perpendicular. This means the angle of intersection, $\angle AOB = 90^\circ$.}

A parallelogram whose diagonals bisect each other at right angles is a rhombus.

So, the shape is a parallelogram with equal and perpendicular diagonals. A quadrilateral with equal diagonals is a rectangle, and a quadrilateral with perpendicular diagonals is a rhombus. A quadrilateral that is both a rectangle and a rhombus is a square.

Alternatively, consider one of the triangles formed by the intersecting diagonals, e.g., $\triangle AOB$. We have $AO = \frac{7}{2}$ cm, $BO = \frac{7}{2}$ cm, and $\angle AOB = 90^\circ$. By the Pythagorean theorem, the side AB (hypotenuse) has length $AB = \sqrt{(\frac{7}{2})^2 + (\frac{7}{2})^2} = \sqrt{\frac{49}{4} + \frac{49}{4}} = \sqrt{\frac{98}{4}} = \sqrt{\frac{49 \times 2}{4}} = \frac{7\sqrt{2}}{2}$ cm. All four sides of the quadrilateral are equal ($\triangle AOB \cong \triangle BOC \cong \triangle COD \cong \triangle DOA$ by SAS). Since the diagonals are equal and perpendicular and bisect each other, the quadrilateral is a square.

Given:

The playground is in the shape of a kite.

Perimeter of the kite = 106 metres.

Length of one side = 23 metres.

To Find:

The lengths of the other three sides.

Solution:

A kite is a quadrilateral that has two distinct pairs of equal-length adjacent sides. Let the lengths of the two distinct sides be $a$ and $b$. The sides of a kite can be represented as having lengths $a, a, b, b$ in sequence around the perimeter.

The perimeter of the kite is the sum of the lengths of its four sides:

Perimeter = $a + a + b + b = 2a + 2b$}

We are given that the perimeter is 106 metres.

Divide the equation by 2:

$a + b = 53$

We are also given that one of the sides is 23 metres. This means that either $a = 23$ or $b = 23$.

Case 1: Let one of the pair of equal sides have length $a = 23$ metres.

Substitute $a = 23$ into the equation $a + b = 53$:

$23 + b = 53$

$b = 53 - 23$

$b = 30$ metres.

In this case, the side lengths of the kite are $a, a, b, b$, which are 23 m, 23 m, 30 m, 30 m.

If one side is 23 m, the other three sides are 23 m, 30 m, and 30 m.

Case 2: Let the other pair of equal sides have length $b = 23$ metres.

Substitute $b = 23$ into the equation $a + b = 53$:

$a + 23 = 53$

$a = 53 - 23$

$a = 30$ metres.

In this case, the side lengths of the kite are $a, a, b, b$, which are 30 m, 30 m, 23 m, 23 m.

If one side is 23 m, the other three sides are 30 m, 30 m, and 23 m.

Both cases result in the same set of side lengths for the kite: two sides of length 23 m and two sides of length 30 m.

Given that one side is 23 metres, the remaining three sides must be one side of length 23 metres and two sides of length 30 metres.

The lengths of the other three sides are 23 metres, 30 metres, and 30 metres.

Given:

READ is a rectangle.

O is the intersection of the diagonals RA and ED.

From the figure, $\angle \text{EDR} = 30^\circ$.

To Find:

$\angle \text{EAR}$, $\angle \text{RAD}$, and $\angle \text{ROD}$.

Solution:

In a rectangle, all interior angles are $90^\circ$. Therefore, $\angle \text{RDA} = 90^\circ$ and $\angle \text{EAD} = 90^\circ$.

Also, the diagonals of a rectangle are equal and bisect each other. Let O be the point of intersection of the diagonals RA and ED. Then $OA = OE = OR = OD$.

The given angle is $\angle \text{EDR} = 30^\circ$. Since O lies on the diagonal ED, the angle $\angle \text{ODR}$ is the same as $\angle \text{EDR}$.

So, $\angle \text{ODR} = 30^\circ$.

Consider $\triangle \text{ODR}$. Since $OD = OR$ (diagonals bisect each other and are equal), $\triangle \text{ODR}$ is an isosceles triangle.

In an isosceles triangle, the angles opposite the equal sides are equal. Thus, $\angle \text{ORD} = \angle \text{ODR} = 30^\circ$.

The sum of angles in a triangle is $180^\circ$. In $\triangle \text{ODR}$, we have:

$\angle \text{ROD} + \angle \text{ORD} + \angle \text{ODR} = 180^\circ$

$\angle \text{ROD} + 30^\circ + 30^\circ = 180^\circ$

$\angle \text{ROD} + 60^\circ = 180^\circ$

$\angle \text{ROD} = 180^\circ - 60^\circ = 120^\circ$.

Now consider the angle at vertex D, which is $\angle \text{RDA} = 90^\circ$. This angle is the sum of $\angle \text{ODR}$ and $\angle \text{ODA}$.

$\angle \text{RDA} = \angle \text{ODR} + \angle \text{ODA}$

$90^\circ = 30^\circ + \angle \text{ODA}$

$\angle \text{ODA} = 90^\circ - 30^\circ = 60^\circ$.

Consider $\triangle \text{OAD}$. Since $OA = OD$, $\triangle \text{OAD}$ is an isosceles triangle.

Therefore, the angles opposite the equal sides are equal: $\angle \text{OAD} = \angle \text{ODA} = 60^\circ$.

The angle $\angle \text{OAD}$ is the same as $\angle \text{RAD}$, as O lies on the diagonal RA.

So, $\angle \text{RAD} = 60^\circ$.

Now consider the angle at vertex A, which is $\angle \text{EAD} = 90^\circ$. This angle is the sum of $\angle \text{EAR}$ and $\angle \text{RAD}$.

$\angle \text{EAD} = \angle \text{EAR} + \angle \text{RAD}$

$90^\circ = \angle \text{EAR} + 60^\circ$

$\angle \text{EAR} = 90^\circ - 60^\circ = 30^\circ$.

Thus, the angles are:

$\angle \text{EAR} = 30^\circ$

$\angle \text{RAD} = 60^\circ$

$\angle \text{ROD} = 120^\circ$

The final answer is $\boxed{\angle EAR = 30^\circ, \angle RAD = 60^\circ, \angle ROD = 120^\circ}$.

Given:

PAIR is a rectangle.

Diagonals PI and AR intersect at M.

$\angle \text{PIA} = 35^\circ$.

To Find:

$\angle \text{ARI}$, $\angle \text{RMI}$ and $\angle \text{PMA}$.

Solution:

In a rectangle, all interior angles are $90^\circ$. Thus, $\angle \text{PAI} = \angle \text{AIR} = \angle \text{IRP} = \angle \text{RPA} = 90^\circ$.

The diagonals of a rectangle are equal and bisect each other. Let M be the intersection of the diagonals AR and PI.

Therefore, $AM = MR = PM = MI$.

Consider $\triangle \text{AMI}$. Since $AM = MI$, $\triangle \text{AMI}$ is an isosceles triangle.

The angles opposite the equal sides are equal: $\angle \text{MAI} = \angle \text{MIA}$.

The given angle $\angle \text{PIA}$ is the same as $\angle \text{MIA}$ (since M lies on the diagonal PI).

So, $\angle \text{MIA} = 35^\circ$.

Therefore, $\angle \text{MAI} = 35^\circ$. This angle is the same as $\angle \text{IAR}$ (since M lies on the diagonal AR).

So, $\angle \text{IAR} = 35^\circ$.

Now consider the right-angled triangle $\triangle \text{AIR}$. The sum of angles in a triangle is $180^\circ$.

$\angle \text{IAR} + \angle \text{AIR} + \angle \text{ARI} = 180^\circ$

$35^\circ + 90^\circ + \angle \text{ARI} = 180^\circ$

$125^\circ + \angle \text{ARI} = 180^\circ$

$\angle \text{ARI} = 180^\circ - 125^\circ$

$\angle \text{ARI} = 55^\circ$.

Next, we find $\angle \text{RMI}$. Consider $\triangle \text{RMI}$. Since $RM = MI$, $\triangle \text{RMI}$ is an isosceles triangle.

The angles opposite the equal sides are equal: $\angle \text{MRI} = \angle \text{MIR}$.

The angle $\angle \text{MIR}$ is part of the right angle $\angle \text{AIR}$.

$\angle \text{MIR} = \angle \text{AIR} - \angle \text{AIP}$. Note that $\angle \text{AIP}$ is the same as $\angle \text{PIA}$.

$\angle \text{MIR} = 90^\circ - 35^\circ = 55^\circ$.

So, $\angle \text{MRI} = 55^\circ$.

In $\triangle \text{RMI}$, the sum of angles is $180^\circ$.

$\angle \text{RMI} + \angle \text{MRI} + \angle \text{MIR} = 180^\circ$

$\angle \text{RMI} + 55^\circ + 55^\circ = 180^\circ$

$\angle \text{RMI} + 110^\circ = 180^\circ$

$\angle \text{RMI} = 180^\circ - 110^\circ$

$\angle \text{RMI} = 70^\circ$.

Finally, we find $\angle \text{PMA}$. $\angle \text{PMA}$ and $\angle \text{RMI}$ are vertically opposite angles formed by the intersection of the diagonals AR and PI.

Vertically opposite angles are equal.

Therefore, $\angle \text{PMA} = \angle \text{RMI} = 70^\circ$.

The required angles are:

$\angle \text{ARI} = 55^\circ$

$\angle \text{RMI} = 70^\circ$

$\angle \text{PMA} = 70^\circ$

The final answer is $\boxed{\angle ARI = 55^\circ, \angle RMI = 70^\circ, \angle PMA = 70^\circ}$.

Given:

ABCD is a parallelogram.

From the figure, $\angle A = 75^\circ$.

To Find:

$\angle B$, $\angle C$, and $\angle D$.

Solution:

In a parallelogram, adjacent angles are supplementary (their sum is $180^\circ$).

Therefore, $\angle A + \angle B = 180^\circ$.

Substituting the given value of $\angle A$:

$75^\circ + \angle B = 180^\circ$

$\angle B = 180^\circ - 75^\circ$

$\angle B = 105^\circ$.

In a parallelogram, opposite angles are equal.

Therefore, $\angle C = \angle A$ and $\angle D = \angle B$.

So, $\angle C = 75^\circ$.

And $\angle D = 105^\circ$.

The required angles are:

$\angle B = 105^\circ$

$\angle C = 75^\circ$

$\angle D = 105^\circ$

The final answer is $\boxed{\angle B = 105^\circ, \angle C = 75^\circ, \angle D = 105^\circ}$.

Given:

PQRS is a parallelogram.

O is the midpoint of diagonal SQ, implying O is the intersection of diagonals.

From the figure, $\angle Q = 100^\circ$, $PS = 6$ cm, and diagonal $SQ = 7.6$ cm.

To Find:

$\angle S$, $\angle R$, length of side PQ, length of side QR, and length of diagonal PR.

Solution:

In a parallelogram, opposite angles are equal.

Therefore, $\angle S = \angle Q$.

Since $\angle Q = 100^\circ$, we have $\angle S = 100^\circ$.

In a parallelogram, adjacent angles are supplementary (sum up to $180^\circ$).

So, $\angle P + \angle Q = 180^\circ$.

$\angle P + 100^\circ = 180^\circ$

$\angle P = 180^\circ - 100^\circ$

$\angle P = 80^\circ$.

Opposite angles are equal, so $\angle R = \angle P$.

Therefore, $\angle R = 80^\circ$.

In a parallelogram, opposite sides are equal.

So, $QR = PS$ and $PQ = RS$.

Since $PS = 6$ cm, we have $QR = 6$ cm.

The given information is $\angle Q = 100^\circ$, $PS = 6$ cm, and $SQ = 7.6$ cm. O is the midpoint of SQ.

The length of side PQ and the length of diagonal PR cannot be uniquely determined using only the provided information ($\angle Q$, $PS$, and $SQ$). A parallelogram is not fully defined by these three parameters alone. Additional information, such as the length of one adjacent side (PQ or RS) or the other diagonal (PR), or an angle related to the diagonal, would be needed to calculate PQ and PR.

The required values that can be determined are:

$\angle S = 100^\circ$

$\angle R = 80^\circ$

QR = 6 cm

PQ and PR cannot be determined from the given information.

The final answer for the calculable values is $\boxed{\angle S = 100^\circ, \angle R = 80^\circ, QR = 6 \text{ cm}}$.

Given:

BEAM is a rhombus.

From the figure, $\angle \text{BMA} = 110^\circ$, which is the angle at vertex M ($\angle M$).

To Find:

$\angle \text{AME}$ and $\angle \text{AEM}$.

Solution:

In a rhombus, opposite angles are equal and adjacent angles are supplementary.

The given angle at vertex M is $\angle M = 110^\circ$.

Since adjacent angles are supplementary, the angle at vertex A (adjacent to M) is:

$\angle A + \angle M = 180^\circ$

$\angle A = 180^\circ - \angle M$

$\angle A = 180^\circ - 110^\circ$

$\angle A = 70^\circ$.

Now consider $\triangle \text{AEM}$. The sides AE and AM are sides of the rhombus.

In a rhombus, all sides are equal, so AE = AM.

This means $\triangle \text{AEM}$ is an isosceles triangle with base EM.

In an isosceles triangle, the angles opposite the equal sides are equal. Therefore, $\angle \text{AEM} = \angle \text{AME}$.

The sum of the interior angles in $\triangle \text{AEM}$ is $180^\circ$. The angles are $\angle \text{EAM}$, $\angle \text{AEM}$, and $\angle \text{AME}$.

The angle $\angle \text{EAM}$ is the vertex angle $\angle A$ of the rhombus.

So, $\angle \text{EAM} = \angle A = 70^\circ$.

Substituting the angles into the sum of angles property for $\triangle \text{AEM}$:

$\angle \text{EAM} + \angle \text{AEM} + \angle \text{AME} = 180^\circ$

$70^\circ + \angle \text{AEM} + \angle \text{AEM} = 180^\circ$ (Since $\angle \text{AEM} = \angle \text{AME}$)

$70^\circ + 2 \angle \text{AEM} = 180^\circ$

$2 \angle \text{AEM} = 180^\circ - 70^\circ$

$2 \angle \text{AEM} = 110^\circ$

$\angle \text{AEM} = \frac{110^\circ}{2}$

$\angle \text{AEM} = 55^\circ$.

Since $\angle \text{AME} = \angle \text{AEM}$, we also have $\angle \text{AME} = 55^\circ$.

The required angles are:

$\angle \text{AME} = 55^\circ$

$\angle \text{AEM} = 55^\circ$

The final answer is $\boxed{\angle AME = 55^\circ, \angle AEM = 55^\circ}$.

Given:

FIST is a parallelogram.

Diagonals FI and ST intersect at O.

From the figure, $\angle \text{TSO} = 50^\circ$ and $\angle \text{OFS} = 40^\circ$.

To Find:

$\angle \text{SFT}$, $\angle \text{OST}$, and $\angle \text{STO}$.

Solution:

In a parallelogram, diagonals bisect each other.

So, O is the midpoint of ST and also the midpoint of FI.

Therefore, $SO = OT$ and $FO = OI$.

Consider $\triangle \text{SOT}$. Since $SO = OT$, $\triangle \text{SOT}$ is an isosceles triangle.

The angles opposite the equal sides are equal. Thus, $\angle \text{OST} = \angle \text{OTS}$.

The given angle $\angle \text{TSO}$ is the same as $\angle \text{OST}$.

So, $\angle \text{OST} = 50^\circ$. (Found one required angle)

Therefore, $\angle \text{OTS} = 50^\circ$. This angle is the same as $\angle \text{STO}$.

So, $\angle \text{STO} = 50^\circ$. (Found another required angle)

In $\triangle \text{SOT}$, the sum of angles is $180^\circ$.

$\angle \text{SOT} + \angle \text{OST} + \angle \text{OTS} = 180^\circ$

$\angle \text{SOT} + 50^\circ + 50^\circ = 180^\circ$

$\angle \text{SOT} + 100^\circ = 180^\circ$

$\angle \text{SOT} = 80^\circ$.

The angles around the intersection point O sum to $360^\circ$. Vertically opposite angles are equal.

$\angle \text{SOT}$ and $\angle \text{FOI}$ are vertically opposite, so $\angle \text{FOI} = \angle \text{SOT} = 80^\circ$.

$\angle \text{SOF}$ and $\angle \text{TOI}$ are vertically opposite, so $\angle \text{SOF} = \angle \text{TOI}$.

The sum of all angles around O is $360^\circ$: $\angle \text{SOT} + \angle \text{SOF} + \angle \text{FOI} + \angle \text{TOI} = 360^\circ$.

$80^\circ + \angle \text{SOF} + 80^\circ + \angle \text{SOF} = 360^\circ$

$160^\circ + 2 \angle \text{SOF} = 360^\circ$

$2 \angle \text{SOF} = 200^\circ$

$\angle \text{SOF} = 100^\circ$.

So, $\angle \text{SOF} = 100^\circ$ and $\angle \text{TOI} = 100^\circ$.

Consider $\triangle \text{SOF}$. We have $\angle \text{SOF} = 100^\circ$. The given angle $\angle \text{OFS}$ is the same as $\angle \text{SFO}$.

So, $\angle \text{SFO} = 40^\circ$.

In $\triangle \text{SOF}$, sum of angles is $180^\circ$:

$\angle \text{SOF} + \angle \text{SFO} + \angle \text{FSO} = 180^\circ$

$100^\circ + 40^\circ + \angle \text{FSO} = 180^\circ$

$140^\circ + \angle \text{FSO} = 180^\circ$

$\angle \text{FSO} = 40^\circ$.

Now, let's find the full angle at vertex S, $\angle \text{FST}$.

$\angle \text{FST} = \angle \text{FSO} + \angle \text{OST} = 40^\circ + 50^\circ = 90^\circ$.

Since $\angle \text{FST} = 90^\circ$ and FIST is a parallelogram, it must be a rectangle.

In a rectangle, all interior angles are $90^\circ$.

Therefore, the angle at vertex F, which is $\angle \text{SFT}$, is $90^\circ$. (Found the third required angle)

The required angles are:

$\angle \text{SFT} = 90^\circ$

$\angle \text{OST} = 50^\circ$

$\angle \text{STO} = 50^\circ$

The final answer is $\boxed{\angle SFT = 90^\circ, \angle OST = 50^\circ, \angle STO = 50^\circ}$.

Given:

YOUR is a parallelogram.

$\angle RUO = 120^\circ$.

OY is extended to point S.

$\angle SRY = 50^\circ$.

To Find:

The value of $\angle YSR$. (The question asks for the value of x, but x is not labelled in the figure. We assume x refers to $\angle YSR$ as this is the angle being calculated in $\triangle YSR$).

Solution:

In parallelogram YOUR, consecutive interior angles are supplementary.

Given $\angle RUO = 120^\circ$. This is the interior angle at vertex U.

The interior angle at vertex Y, $\angle OYR$, is consecutive to $\angle RUO$.

$\angle OYR + \angle RUO = 180^\circ$

$\angle OYR + 120^\circ = 180^\circ$

$\angle OYR = 180^\circ - 120^\circ = 60^\circ$.

In a parallelogram, opposite angles are equal. So, $\angle Y = \angle U = 120^\circ$ and $\angle O = \angle R = 60^\circ$. The angle at vertex Y, $\angle OYR$, is indeed $\angle Y = 120^\circ$.

Since OY is extended to S, the points O, Y, and S are collinear. Therefore, $\angle OYR$ and $\angle SYR$ form a linear pair.

$\angle OYR + \angle SYR = 180^\circ$

$120^\circ + \angle SYR = 180^\circ$

$\angle SYR = 180^\circ - 120^\circ = 60^\circ$

Now consider the triangle $\triangle$YSR. The sum of the interior angles of a triangle is $180^\circ$.

The angles in $\triangle$YSR are $\angle YSR$, $\angle SYR$, and $\angle SRY$.

We know $\angle SYR = 60^\circ$ and we are given $\angle SRY = 50^\circ$.

$\angle YSR + \angle SYR + \angle SRY = 180^\circ$

$\angle YSR + 60^\circ + 50^\circ = 180^\circ$

$\angle YSR + 110^\circ = 180^\circ$

$\angle YSR = 180^\circ - 110^\circ$

$\angle YSR = 70^\circ$

If 'x' in the question refers to $\angle YSR$, then $x = 70^\circ$.

Given:

Kite WEAR with $\angle E = 70^\circ$ and $\angle R = 80^\circ$.

To Find:

The measures of $\angle W$ and $\angle A$.

Solution:

In a kite, one pair of opposite angles is equal. Given $\angle E = 70^\circ$ and $\angle R = 80^\circ$, these angles are at adjacent vertices based on the standard vertex naming WEAR. Assuming the equal opposite angles are $\angle W$ and $\angle A$.

Let $\angle W = \angle A = k$.

The sum of the interior angles of a quadrilateral is $360^\circ$.

$\angle W + \angle E + \angle A + \angle R = 360^\circ$

$k + 70^\circ + k + 80^\circ = 360^\circ$

$2k + 150^\circ = 360^\circ$

$2k = 360^\circ - 150^\circ$

$2k = 210^\circ$

$k = \frac{210^\circ}{2}$

$k = 105^\circ$

So, $\angle W = 105^\circ$ and $\angle A = 105^\circ$.

The remaining two angles are $\angle W = 105^\circ$ and $\angle A = 105^\circ$.

Given:

A rectangle MORE. Diagonals MR and OE intersect at Y. Point X is on side RE.

To Answer:

The given questions (i) to (v) by providing appropriate reasons.

Solution:

(i) Is RE = OM?

Yes.

Reason: In a rectangle, opposite sides are equal in length. RE and OM are opposite sides of rectangle MORE.

(ii) Is ∠MYO = ∠RXE?

No.

Reason: Y is the intersection of the diagonals MR and OE. ∠MYO is an angle formed by the intersecting diagonals at point Y. Point X is on the side RE. Since X lies on the line segment RE, the points R, X, and E are collinear. If X is strictly between R and E, the angle ∠RXE, formed by these collinear points, is a straight angle measuring $180^\circ$. The angle ∠MYO, which is an angle between the diagonals of a rectangle, is typically an acute or obtuse angle (equal to $90^\circ$ only if the rectangle is a square) but is always less than $180^\circ$. Therefore, ∠MYO cannot be equal to ∠RXE.

(iii) Is ∠MOY = ∠REX?

No.

Reason: In triangle MYO, ∠MOY is the angle at vertex O, formed by the side MO and the segment OY (part of the diagonal OE). Point X is on the side RE. Since X lies on the line segment RE, the points R, X, and E are collinear. The angle ∠REX is the angle at vertex E in triangle RXE. However, if R, X, and E are collinear, they do not form a non-degenerate triangle, and the interior angles are not defined in the usual sense. If interpreted as an angle on the straight line RE, ∠REX would be $0^\circ$ or $180^\circ$. ∠MOY is an angle in the non-degenerate triangle MYO, and is not equal to $0^\circ$ or $180^\circ$. Therefore, ∠MOY is not equal to ∠REX.

(iv) Is ∆MYO ≅ ∆RXE?

No.

Reason: Y is the intersection of the diagonals of the rectangle, and M, O are vertices. ΔMYO is a non-degenerate triangle formed by vertices M, Y, and O. Point X is on the side RE. Since X lies on the line segment RE, the points R, X, and E are collinear. Collinear points R, X, and E form a degenerate triangle with zero area. A non-degenerate triangle (ΔMYO) cannot be congruent to a degenerate triangle (ΔRXE).

(v) Is MY = RX?

No.

Reason: MY is the length of half of the diagonal MR (since Y is the midpoint of the diagonal). RX is the length of a segment on the side RE. In a general rectangle, the length of a diagonal is different from the length of a side (unless it is a square). Therefore, half of a diagonal is not generally equal to a part of a side. MY = $\frac{1}{2} \$ MR, and RX is a segment of RE. There is no general property that states MY is equal to RX for any point X on RE in a rectangle. Equality would hold only in specific circumstances not given in the problem.

Given:

LOST is a parallelogram with SN⊥OL and SM⊥LT.

To Find:

∠STM, ∠SON and ∠NSM.

Solution:

In parallelogram LOST, opposite sides are parallel, and consecutive angles are supplementary. OL || ST and OT || LS.

∠STM is the angle between side ST and the extension of side LT (which is LM). This is the exterior angle at vertex T.

The exterior angle at a vertex of a parallelogram is equal to the interior opposite angle.

The interior opposite angle to ∠T is ∠O.

Therefore, ∠STM = ∠O.

∠SON is the interior angle at vertex O of the parallelogram LOST. ON is the extension of OL.

Therefore, ∠SON = ∠O.

Consider the quadrilateral LNSM. We are given SN⊥OL and SM⊥LT. Since N is on the extension of OL, ∠LNS = $90^\circ$. Since M is on the extension of LT, ∠SML = $90^\circ$.

The sum of angles in quadrilateral LNSM is $360^\circ$.

∠L + ∠LNS + ∠NSM + ∠SML = $360^\circ$

∠L + $90^\circ$ + ∠NSM + $90^\circ$ = $360^\circ$

∠L + ∠NSM = $180^\circ$

In parallelogram LOST, consecutive angles are supplementary, so ∠L + ∠O = $180^\circ$.

Comparing the two equations, we get ∠NSM = ∠O.

Thus, we have found:

∠STM = ∠O

∠SON = ∠O

∠NSM = ∠O

All three angles are equal to the interior angle ∠O of the parallelogram.

The final answer is $\boxed{\angle STM = \angle SON = \angle NSM = \angle O}$.

Given:

HARE is a trapezium. Based on the diagram, we assume HA is parallel to RE (HA || RE). EP is the bisector of ∠HER, and RP is the bisector of ∠ARE. P is the intersection of EP and RP.

To Find:

∠HAR and ∠EHA.

Solution:

In a trapezium HARE with HA || RE, the sum of consecutive interior angles on the non-parallel sides (legs) is $180^\circ$. The legs are HE and AR.

So, for leg HE:

Rearranging this equation to find ∠EHA:

Similarly, for leg AR:

Rearranging this equation to find ∠HAR:

We are given that EP bisects ∠HER and RP bisects ∠ARE.

Therefore, $\angle PER = \frac{1}{2}\angle HER$ and $\angle PRE = \frac{1}{2}\angle ARE$.

Consider $\triangle EPR$. The sum of angles in a triangle is $180^\circ$.

$\angle EPR + \angle PER + \angle PRE = 180^\circ$

$\angle EPR + \frac{1}{2}\angle HER + \frac{1}{2}\angle ARE = 180^\circ$

$\angle EPR + \frac{1}{2}(\angle HER + \angle ARE) = 180^\circ$

Since ∠HER + ∠ARE = $180^\circ$ (consecutive interior angles on leg ER between parallel lines HA and RE), we substitute this value:

$\angle EPR + \frac{1}{2}(180^\circ) = 180^\circ$

$\angle EPR + 90^\circ = 180^\circ$

$\angle EPR = 90^\circ$

This shows that the bisectors of the angles on a non-parallel side of a trapezium are perpendicular.

However, the question asks for ∠HAR and ∠EHA. From equations (1) and (2), we have found expressions for these angles in terms of the angles ∠HER and ∠ARE.

Without more information or specific values for angles ∠HER or ∠ARE, we cannot find unique numerical values for ∠HAR and ∠EHA. The answer provides the expressions for these angles based on the properties of the trapezium.

The final answer is $\boxed{\angle HAR = 180^\circ - \angle ARE \text{ and } \angle EHA = 180^\circ - \angle HER}$.

Given:

MODE is a parallelogram.

MQ is the bisector of ∠M.

OQ is the bisector of ∠O.

The bisectors MQ and OQ meet at point Q.

To Find:

The measure of ∠MQO.

Solution:

In a parallelogram, consecutive angles are supplementary.

This means that the sum of adjacent angles is $180^\circ$.

Therefore, for parallelogram MODE, we have:

$\angle M + \angle O = 180^\circ$

Since MQ is the bisector of ∠M, it divides the angle into two equal parts:

$\angle QMO = \frac{1}{2} \angle M$

Since OQ is the bisector of ∠O, it divides the angle into two equal parts:

$\angle QOM = \frac{1}{2} \angle O$

Now consider the triangle $\triangle$MQO.

The sum of the angles in any triangle is $180^\circ$.

In $\triangle$MQO, the angles are $\angle MQO$, $\angle QMO$, and $\angle QOM$.

So, $\angle MQO + \angle QMO + \angle QOM = 180^\circ$

Substitute the expressions for $\angle QMO$ and $\angle QOM$:

$\angle MQO + \frac{1}{2} \angle M + \frac{1}{2} \angle O = 180^\circ$

Factor out $\frac{1}{2}$:

$\angle MQO + \frac{1}{2} (\angle M + \angle O) = 180^\circ$

We know that $\angle M + \angle O = 180^\circ$. Substitute this value into the equation:

$\angle MQO + \frac{1}{2} (180^\circ) = 180^\circ$

$\angle MQO + 90^\circ = 180^\circ$

Subtract $90^\circ$ from both sides:

$\angle MQO = 180^\circ - 90^\circ$

$\angle MQO = 90^\circ$

Therefore, the measure of ∠MQO is $90^\circ$.

Given:

ATEF is a rectangle.

Points F and B are given such that EF = EB.

From the figure, $\angle TFB = x$ and $\angle FEB = y$.

To Find:

The values of x and y.

Solution:

Since ATEF is a rectangle, all its interior angles are $90^\circ$.

So, $\angle TFE = 90^\circ$.

From the figure, it appears that $\angle TFB$ is a part of $\angle TFE$. Therefore, $\angle TFB + \angle EFB = \angle TFE$.

$x + \angle EFB = 90^\circ$

This implies $\angle EFB = 90^\circ - x$.

We are given that EF = EB.

Consider $\triangle$EFB. Since two sides EF and EB are equal, $\triangle$EFB is an isosceles triangle.

In an isosceles triangle, the angles opposite the equal sides are equal.

So, the angle opposite side EF ($\angle EBF$) is equal to the angle opposite side EB ($\angle EFB$).

$\angle EBF = \angle EFB = 90^\circ - x$.

The sum of the interior angles in any triangle is $180^\circ$.

In $\triangle$EFB, the angles are $\angle FEB$, $\angle EFB$, and $\angle EBF$.

$\angle FEB + \angle EFB + \angle EBF = 180^\circ$

Substitute the given and derived angle measures:

$y + (90^\circ - x) + (90^\circ - x) = 180^\circ$

$y + 180^\circ - 2x = 180^\circ$

Subtract $180^\circ$ from both sides:

$y - 2x = 0$

$y = 2x$

From the figure, the angle $\angle FEB = y$ appears to be a right angle. If $\triangle$EFB is a right-angled isosceles triangle with EF = EB, the right angle must be at E ($\angle FEB$). This would make the other two angles equal to $45^\circ$. Let's assume $\angle FEB = 90^\circ$ based on the appearance of the figure and common problem types.

Assume $y = 90^\circ$.

Substitute this value into the equation $y = 2x$:

$90^\circ = 2x$

Divide by 2:

$x = \frac{90^\circ}{2}$

$x = 45^\circ$

Let's check if these values are consistent with all conditions:

If $x = 45^\circ$ and $y = 90^\circ$:

$\angle EFB = 90^\circ - x = 90^\circ - 45^\circ = 45^\circ$.

$\angle EBF = 90^\circ - x = 90^\circ - 45^\circ = 45^\circ$.

In $\triangle$EFB, angles are $\angle FEB = y = 90^\circ$, $\angle EFB = 45^\circ$, $\angle EBF = 45^\circ$. The sum is $90^\circ + 45^\circ + 45^\circ = 180^\circ$. This is a valid triangle.

Also, $\angle EFB = \angle EBF = 45^\circ$, which is consistent with EF=EB in $\triangle$EFB.

Finally, check the angle at F: $\angle TFB = x = 45^\circ$. $\angle EFB = 45^\circ$. $\angle TFB + \angle EFB = 45^\circ + 45^\circ = 90^\circ$. This matches $\angle TFE = 90^\circ$ from the rectangle property and the assumption that B's position results in $\angle TFB$ and $\angle EFB$ being adjacent parts of $\angle TFE$.

The values of x and y are $x = 45^\circ$ and $y = 90^\circ$.

Given:

From the figure, ABDH is a parallelogram.

CEFG is a parallelogram.

The angles are labelled as follows:

$\angle DAB = 120^\circ$

$\angle ECG = x$

$\angle CEG = 60^\circ$

To Find:

The value of x.

Solution:

In parallelogram ABDH, opposite angles are equal.

So, $\angle DHB = \angle DAB = 120^\circ$.

Consecutive angles in a parallelogram are supplementary.

So, $\angle ADB + \angle DAB = 180^\circ$.

$\angle ADB + 120^\circ = 180^\circ$

$\angle ADB = 180^\circ - 120^\circ = 60^\circ$.

Also, $\angle HBD + \angle BDH = 180^\circ$.

$\angle H + \angle A = 180^\circ$. $\angle H = 180^\circ - 120^\circ = 60^\circ$.

$\angle ABD = \angle AHD = 60^\circ$.

In parallelogram CEFG, opposite angles are equal.

So, $\angle CFG = \angle CEG = 60^\circ$.

Consecutive angles in a parallelogram are supplementary.

So, $\angle ECG + \angle CEG = 180^\circ$ (angles on leg CG and CE, incorrect as these are adjacent vertices, not necessarily on the same leg between parallel sides unless it's a specific orientation).

Consecutive angles are at adjacent vertices. $\angle GCE + \angle CEG = 180^\circ$ is not guaranteed unless GC || FE. In parallelogram CEFG, CE || GF and CG || EF.

Adjacent angles are $\angle GCE$ and $\angle CEG$. Their sum is $180^\circ$ only if GC || FE which is true. Wait, adjacent angles in a parallelogram like CEFG are $\angle C$ and $\angle E$, $\angle E$ and $\angle F$, etc. So $\angle GCE + \angle CEG = 180^\circ$ is not true in general unless it refers to a specific pair.

Let's use opposite angles directly. $\angle GCE$ is the angle at vertex C, which is $\angle ECG = x$.

In parallelogram CEFG, $\angle F = \angle C$ and $\angle E = \angle G$.

So $\angle CFG = \angle CEG = 60^\circ$ is incorrect. The opposite angles are $\angle C = \angle F$ and $\angle E = \angle G$.

So, $\angle CFG = \angle C$ (not $\angle CEG$). $\angle CEG = 60^\circ$ is given as the angle at vertex E.

In parallelogram CEFG, opposite angles are equal.

$\angle C = \angle F$ (i.e., $\angle ECG = \angle CFG$)

$\angle E = \angle G$ (i.e., $\angle CEG = \angle CGF$)

Given $\angle CEG = 60^\circ$. So $\angle G = 60^\circ$.

Consecutive angles are supplementary:

$\angle C + \angle E = 180^\circ$

$\angle ECG + \angle CEG = 180^\circ$

$x + 60^\circ = 180^\circ$

Subtract $60^\circ$ from both sides:

$x = 180^\circ - 60^\circ$

$x = 120^\circ$

Let's verify. If $x=120^\circ$, then $\angle C = 120^\circ$. $\angle E=60^\circ$. $\angle C + \angle E = 120 + 60 = 180^\circ$ (supplementary). Opposite angles $\angle C = \angle F = 120^\circ$, $\angle E = \angle G = 60^\circ$. Sum of angles = $120+60+120+60=360^\circ$. This is consistent for parallelogram CEFG.

The information from parallelogram ABDH ($\angle DAB = 120^\circ$) is consistent with the calculated value of x=120 for $\angle ECG$. It seems the problem implies that the angle at A in the first parallelogram is equal to the angle at C in the second parallelogram. This might be intended visually from the figure, suggesting a relationship between the two parallelograms, but it is not explicitly stated that $\angle DAB = \angle ECG$ or that they are corresponding parts of the "ship" structure that must have equal angles.

However, finding x only requires the properties of parallelogram CEFG.

The value of x is $120^\circ$.

Given:

ABCD is a rhombus.

Semicircles are drawn on each side of rhombus ABCD.

A grid is provided in the figure.

To Find:

The radius of the semicircle drawn on each side of rhombus ABCD.

Solution:

The semicircles are drawn on each side of the rhombus ABCD.

This means that each side of the rhombus acts as the diameter of the semicircle drawn on that side.

To find the radius of the semicircle, we first need to find the length of the side of the rhombus.

We can determine the side length by using the grid provided in the figure.

Let's consider the side AB of the rhombus. By observing the grid, we can see the horizontal and vertical displacement between vertices A and B.

Starting from A, to reach B, we move 3 units horizontally (to the right) and 1 unit vertically (downwards).

Using the Pythagorean theorem, the length of the side AB is the hypotenuse of a right triangle with legs of length 3 units and 1 unit.

Length of side AB = $\sqrt{(3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10}$ units.

Since ABCD is a rhombus, all its sides are equal in length.

So, the length of each side of the rhombus ABCD is $\sqrt{10}$ units.

The diameter of each semicircle is equal to the side length of the rhombus.

Diameter = $\sqrt{10}$ units.

The radius of a circle (or a semicircle) is half of its diameter.

Radius = $\frac{\text{Diameter}}{2} = \frac{\sqrt{10}}{2}$ units.

(Here, 'units' refers to the spacing of the grid lines).

The radius of the semicircle drawn on each side of rhombus ABCD is $\frac{\sqrt{10}}{2}$.

Given:

ABCDE is a regular pentagon.

AM is the bisector of $\angle A$, meeting the side CD at M.

To Find:

The measure of $\angle AMC$.

Solution:

In a regular pentagon, all interior angles are equal. The measure of each interior angle is given by the formula $\frac{(n-2) \times 180^\circ}{n}$, where $n=5$.

Measure of each interior angle $= \frac{(5-2) \times 180^\circ}{5} = \frac{3 \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ$.

So, $\angle A = \angle B = \angle C = \angle D = \angle E = 108^\circ$.

All sides of a regular pentagon are equal: AB = BC = CD = DE = EA.

AM is the bisector of $\angle A$.

So, $\angle EAM = \angle BAM = \frac{1}{2} \angle A = \frac{1}{2} \times 108^\circ = 54^\circ$.

Consider the triangle $\triangle$ABC.

AB = BC (sides of a regular pentagon).

$\triangle$ABC is an isosceles triangle.

$\angle ABC = 108^\circ$.

The base angles are equal: $\angle BAC = \angle BCA = \frac{180^\circ - 108^\circ}{2} = \frac{72^\circ}{2} = 36^\circ$.

Now consider the angle $\angle CAD$.

$\angle CAD = \angle A - \angle BAC - \angle EAD$.

By symmetry of a regular pentagon, $\triangle$ABC, $\triangle$BCD, $\triangle$CDE, $\triangle$DEA, $\triangle$EAB are congruent isosceles triangles.

So, $\angle BAC = \angle CAD = \angle DAE = 36^\circ$. (Alternatively, the angle at the center divided by 5 is the angle subtended by each side at the center, which is $360/5=72^\circ$. The angle at the vertex is related to this. The angle subtended by a side at the center is $72^\circ$. In the isosceles triangle formed by two vertices and the center, the base angles are $(180-72)/2 = 54^\circ$. So the angle at the vertex of the pentagon is $2 \times 54^\circ = 108^\circ$. The diagonal AC forms an isosceles triangle ABC. Angle ABC=108. Angles BAC and BCA are (180-108)/2 = 36. Similarly, in triangle CDE, angle CDE=108, angles DCE and DEC are 36. In triangle DEA, angle DEA=108, angles EDA and EAD are 36. Angle EAD=36. Angle BAC=36. So angle CAD = Angle A - Angle BAC - Angle EAD = 108 - 36 - 36 = 36).

So, $\angle CAD = 36^\circ$.

AM is the bisector of $\angle A$. $\angle BAM = 54^\circ$.

$\angle CAM = \angle BAM - \angle BAC = 54^\circ - 36^\circ = 18^\circ$.

We need $\angle AMC$. Consider $\triangle$AMC.

We know $\angle CAM = 18^\circ$.

We know $\angle C = \angle BCD = 108^\circ$.

We need to find $\angle ACM$. $\angle ACM$ is part of $\angle BCD$.

$\angle BCD = \angle BCA + \angle ACD$.

We found $\angle BCA = 36^\circ$.

Consider $\triangle$BCD. BC = CD, so it is isosceles. $\angle CBD = \angle CDB = (180-108)/2 = 36^\circ$.

$\angle ACD = \angle BCD - \angle BCA = 108^\circ - 36^\circ = 72^\circ$.

So, $\angle ACM = \angle ACD = 72^\circ$.

In $\triangle$AMC, the angles are $\angle CAM = 18^\circ$, $\angle ACM = 72^\circ$, and $\angle AMC$.

Sum of angles in $\triangle$AMC $= 180^\circ$.

$\angle CAM + \angle ACM + \angle AMC = 180^\circ$

$18^\circ + 72^\circ + \angle AMC = 180^\circ$

$90^\circ + \angle AMC = 180^\circ$

$\angle AMC = 180^\circ - 90^\circ$

$\angle AMC = 90^\circ$

The measure of $\angle AMC$ is $90^\circ$.

Given:

Quadrilateral EFGH is a rectangle.

J is the point of intersection of the diagonals EG and FH.

Length of segment JF = $8x + 4$.

Length of diagonal EG = $24x - 8$.

To Find:

The value of x.

Solution:

We know that the diagonals of a rectangle are equal in length and bisect each other.

Let the diagonals be EG and FH. Since J is the point of intersection, J is the midpoint of both diagonals.

Therefore, the segments from the vertices to the intersection point are equal:

$EJ = JG = FJ = JH$

This implies that the length of segment FJ is half the length of the diagonal EG.

$FJ = \frac{1}{2} \times EG$

Substitute the given expressions for FJ and EG:

$8x + 4 = \frac{1}{2} (24x - 8)$

Multiply both sides by 2 to eliminate the fraction:

$2(8x + 4) = 24x - 8$

Distribute the 2 on the left side:

$16x + 8 = 24x - 8$

Subtract $16x$ from both sides:

$8 = 24x - 16x - 8$

$8 = 8x - 8$

Add 8 to both sides:

$8 + 8 = 8x$

$16 = 8x$

Divide both sides by 8:

$\frac{16}{8} = x$

$x = 2$

Thus, the value of x is 2.

Verification (Optional):

Substitute $x=2$ into the expressions for FJ and EG:

$FJ = 8(2) + 4 = 16 + 4 = 20$

$EG = 24(2) - 8 = 48 - 8 = 40$

Check if FJ is half of EG:

$20 = \frac{1}{2} \times 40$

$20 = 20$

The values match, confirming the value of x is correct.

Given:

The figure shows a parallelogram with diagonals intersecting at point O.

The lengths of the segments of the diagonals are given as:

AO = $y+7$

OC = $20$

BO = $x+y$

OD = $16$

To Find:

The values of x and y.

Solution:

We know that the diagonals of a parallelogram bisect each other.

This means the point of intersection O is the midpoint of both diagonals AC and BD.

Therefore, the segments of each diagonal are equal in length:

Length of AO = Length of OC

$y+7 = 20$

Length of BO = Length of OD

$x+y = 16$

Now we solve these equations for x and y.

From the first equation:

$y + 7 = 20$

Subtract 7 from both sides:

$y = 20 - 7$

$y = 13$

Substitute the value of y ($y=13$) into the second equation:

$x + y = 16$

$x + 13 = 16$

Subtract 13 from both sides:

$x = 16 - 13$

$x = 3$

The values of x and y are $x=3$ and $y=13$.

Given:

The figure shows a kite with side lengths $3x+2$, $y$, $5x-1$, and $3x+2$. One angle is $120^\circ$.

To Find:

The values of x and y.

Solution:

In a kite, there are two distinct pairs of equal-length adjacent sides.

From the figure, the side lengths are $3x+2, y, 5x-1, 3x+2$. The pair of equal adjacent sides must involve the repeated length $3x+2$.

Let the pairs of equal adjacent sides be $(3x+2, y)$ and $(3x+2, 5x-1)$. This implies the side of length $3x+2$ is adjacent to sides of length $y$ and $5x-1$. This occurs at one vertex.

Thus, the other two sides must be equal in length:

$y = 5x - 1$

Also, the sides adjacent to the angle $120^\circ$ must be involved. The figure suggests the sides of length $3x+2$ meet at a vertex and form a pair of equal adjacent sides. If this is the case, the other pair of equal adjacent sides would have lengths $y$ and $5x-1$, meaning $y = 5x-1$. The angles at the vertices where unequal sides meet are equal (in this case, the angles opposite the axis of symmetry formed by the diagonal between the vertices with equal adjacent sides). The angle $120^\circ$ is likely one of these equal angles.

However, the presence of $3x+2$ appearing twice as non-adjacent sides in the sequence $3x+2, y, 5x-1, 3x+2$ in the figure strongly suggests that the intended equal adjacent sides are $(3x+2, 3x+2)$ and $(y, 5x-1)$.

So, $y = 5x - 1$ and $3x+2 = 3x+2$ (which is always true).

If the sides are $3x+2$, $y$, $5x-1$, $3x+2$, and it's a kite, the pairs of equal adjacent sides are $(3x+2, y)$ and $(5x-1, 3x+2)$. This implies $3x+2 = 5x-1$ and $y = 3x+2$. (Or $3x+2=3x+2$ and $y=5x-1$). Based on the visual representation, the sides labelled $3x+2$ appear equal and adjacent. The sides labelled $y$ and $5x-1$ appear equal and adjacent.

Assuming the pairs of equal adjacent sides are $(3x+2, y)$ and $(3x+2, 5x-1)$, or $(y, 5x-1)$ and $(3x+2, 3x+2)$. The second case leads to all sides being equal, i.e., a rhombus.

Let's assume the intended equal adjacent sides are those that yield a solvable system from the side expressions alone. The most likely scenario for a basic problem is where the lengths of the two distinct pairs of equal adjacent sides are given by the expressions.

Assuming the side lengths are $3x+2, y, 5x-1$. The kite has two pairs of equal adjacent sides. Let these be $3x+2 = y$ and $3x+2 = 5x-1$. (This makes it a rhombus, which is a kite).

From the equality $3x + 2 = 5x - 1$:

$2 + 1 = 5x - 3x$

$3 = 2x$

$x = \frac{3}{2} = 1.5$

From the equality $y = 3x + 2$:

$y = 3(1.5) + 2$

$y = 4.5 + 2$

$y = 6.5$

This gives side lengths $6.5, 6.5, 6.5, 6.5$. This is a rhombus, a type of kite. The angle $120^\circ$ is consistent with a rhombus (where angles are $120^\circ, 60^\circ, 120^\circ, 60^\circ$).

This interpretation fits the criteria of a kite and leads to a unique solution.

The values of x and y are $x = 1.5$ and $y = 6.5$.

Given:

The figure shows a trapezium ABCD, where AB || DC.

The interior angles are given as:

$\angle A = 110^\circ$

$\angle B = x$

$\angle C = 80^\circ$

$\angle D = 90^\circ$

To Find:

The value of x.

Solution:

In a trapezium, the sum of the interior angles on the same non-parallel side (leg) is $180^\circ$ when the other two sides are parallel.

Given that AB || DC, the segment BC is a transversal intersecting the parallel lines AB and DC. The angles $\angle B$ and $\angle C$ are consecutive interior angles on the leg BC.

Therefore, the sum of $\angle B$ and $\angle C$ is $180^\circ$.

$\angle B + \angle C = 180^\circ$

Substitute the given values for $\angle B$ and $\angle C$:

$x + 80^\circ = 180^\circ$

To find the value of x, subtract $80^\circ$ from both sides of the equation:

$x = 180^\circ - 80^\circ$

$x = 100^\circ$

The value of x is $100^\circ$.

Note: Using the same property for the leg AD, we would expect $\angle A + \angle D = 180^\circ$. However, $110^\circ + 90^\circ = 200^\circ$, which is not $180^\circ$. This indicates an inconsistency in the given angle measures for angles A and D relative to the parallel sides. However, the value of x can be determined solely from the properties related to the leg BC.

Given:

A quadrilateral has two angles, each measuring $75^\circ$.

The other two angles are equal in measure.

To Find:

1. The measure of the two equal angles.

2. The possible figures (quadrilaterals) that satisfy these conditions.

Solution:

Let the two angles of the quadrilateral be $75^\circ$ and $75^\circ$.

Let the measure of each of the other two equal angles be $x^\circ$.

The sum of the interior angles of a quadrilateral is $360^\circ$.

Therefore, we can write the equation:

$75^\circ + 75^\circ + x + x = 360^\circ$

Combine the constant terms and the terms with x:

$150^\circ + 2x = 360^\circ$

Subtract $150^\circ$ from both sides of the equation:

$2x = 360^\circ - 150^\circ$

$2x = 210^\circ$

Divide both sides by 2:

$x = \frac{210^\circ}{2}$

$x = 105^\circ$

So, the measure of each of the other two equal angles is $105^\circ$.

Possible Figures:

The angles of the quadrilateral are $75^\circ, 75^\circ, 105^\circ, 105^\circ$.

Let's consider the possible arrangements of these angles.

Case 1: The two $75^\circ$ angles are opposite to each other.

If two opposite angles are $75^\circ$, then the other two opposite angles must be $105^\circ$ each (as calculated). In a quadrilateral, if opposite angles are equal, it is a parallelogram.

The angles would be arranged as $75^\circ, 105^\circ, 75^\circ, 105^\circ$ (in order around the quadrilateral). In this case, adjacent angles sum to $75^\circ + 105^\circ = 180^\circ$, which is a property of parallelograms. Thus, a parallelogram is a possible figure.

Case 2: The two $75^\circ$ angles are adjacent to each other.

If two adjacent angles are $75^\circ$, then the angles are arranged as $75^\circ, 75^\circ, 105^\circ, 105^\circ$ (in order around the quadrilateral). Let the angles be $\angle A=75^\circ, \angle B=75^\circ, \angle C=105^\circ, \angle D=105^\circ$.

In this case, $\angle A + \angle D = 75^\circ + 105^\circ = 180^\circ$. This implies that the side AB is parallel to the side DC.

Also, $\angle B + \angle C = 75^\circ + 105^\circ = 180^\circ$. This implies that the side AB is parallel to the side DC.

Since AB || DC, the quadrilateral is a trapezoid. Specifically, because the non-parallel sides AD and BC connect angles that sum to $180^\circ$ with the parallel sides, and the adjacent angles at each base are equal ($75^\circ, 75^\circ$ and $105^\circ, 105^\circ$), this is an isosceles trapezoid.

Thus, an isosceles trapezoid is also a possible figure.

The measure of the two equal angles is $105^\circ$.

The possible figures are a parallelogram or an isosceles trapezoid.

Given:

In quadrilateral PQRS, the measures of three angles are:

$\angle P = 50^\circ$

$\angle Q = 50^\circ$

$\angle R = 60^\circ$

To Find:

1. The measure of $\angle S$.

2. Whether the quadrilateral is convex or concave.

Solution:

The sum of the interior angles of a quadrilateral is $360^\circ$.

Therefore, for quadrilateral PQRS:

$\angle P + \angle Q + \angle R + \angle S = 360^\circ$

Substitute the given values of the angles:

$50^\circ + 50^\circ + 60^\circ + \angle S = 360^\circ$

Add the known angles:

$160^\circ + \angle S = 360^\circ$

Subtract $160^\circ$ from both sides to find $\angle S$:

$\angle S = 360^\circ - 160^\circ$

$\angle S = 200^\circ$

The measure of $\angle S$ is $200^\circ$.

Convex or Concave:

A quadrilateral is considered convex if all its interior angles are less than $180^\circ$.

A quadrilateral is considered concave if at least one of its interior angles is greater than $180^\circ$.

The angles of quadrilateral PQRS are $50^\circ, 50^\circ, 60^\circ$, and $200^\circ$.

Since $\angle S = 200^\circ$, which is greater than $180^\circ$, the quadrilateral has one interior angle greater than $180^\circ$.

Therefore, the quadrilateral PQRS is concave.

Given:

A quadrilateral in which:

1. Both pairs of opposite angles are equal.

2. Both pairs of opposite angles are supplementary (sum up to $180^\circ$).

To Find:

The measure of each angle of the quadrilateral.

Solution:

Let the quadrilateral be ABCD.

Let its angles be $\angle A, \angle B, \angle C,$ and $\angle D$.

According to the given information, the opposite angles are equal:

$\angle A = \angle C$

$\angle B = \angle D$

Also, the opposite angles are supplementary. This means:

$\angle A + \angle C = 180^\circ$

$\angle B + \angle D = 180^\circ$

Let's use the equality property in the supplementary property.

For the first pair of opposite angles ($\angle A$ and $\angle C$):

Since $\angle A = \angle C$ and $\angle A + \angle C = 180^\circ$, we can substitute $\angle A$ for $\angle C$ in the second equation:

$\angle A + \angle A = 180^\circ$

$2 \angle A = 180^\circ$

Divide by 2:

$\angle A = \frac{180^\circ}{2}$

$\angle A = 90^\circ$

Since $\angle A = \angle C$, we have:

$\angle C = 90^\circ$

Now, for the second pair of opposite angles ($\angle B$ and $\angle D$):

Since $\angle B = \angle D$ and $\angle B + \angle D = 180^\circ$, we can substitute $\angle B$ for $\angle D$ in the second equation:

$\angle B + \angle B = 180^\circ$

$2 \angle B = 180^\circ$

Divide by 2:

$\angle B = \frac{180^\circ}{2}$

$\angle B = 90^\circ$

Since $\angle B = \angle D$, we have:

$\angle D = 90^\circ$

Thus, all four angles of the quadrilateral are $90^\circ$.

The measure of each angle is $90^\circ$.

Given:

A regular octagon.

To Find:

The measure of each interior angle of the regular octagon.

Solution:

A regular octagon is a polygon with 8 equal sides and 8 equal interior angles.

The formula for the sum of the interior angles of a polygon with $n$ sides is given by:

Sum of interior angles $= (n-2) \times 180^\circ$

For an octagon, the number of sides is $n = 8$.

So, the sum of the interior angles of an octagon is:

Sum $= (8-2) \times 180^\circ$

Sum $= 6 \times 180^\circ$

Sum $= 1080^\circ$

Since the octagon is regular, all its interior angles are equal.

To find the measure of each interior angle, we divide the sum of the interior angles by the number of sides (or angles), which is 8.

Measure of each interior angle $= \frac{\text{Sum of interior angles}}{\text{Number of sides (n)}}$

Measure of each interior angle $= \frac{1080^\circ}{8}$

Measure of each interior angle $= 135^\circ$

Therefore, the measure of each angle of a regular octagon is $135^\circ$.

Alternate Method using Exterior Angles:

The sum of the exterior angles of any convex polygon is $360^\circ$.

For a regular n-sided polygon, each exterior angle is equal.

Measure of each exterior angle $= \frac{360^\circ}{n}$

For a regular octagon, $n = 8$.

Measure of each exterior angle $= \frac{360^\circ}{8}$

Measure of each exterior angle $= 45^\circ$

The interior angle and the adjacent exterior angle at any vertex of a polygon are supplementary (they form a straight line, summing to $180^\circ$).

Measure of each interior angle + Measure of each exterior angle $= 180^\circ$

Measure of each interior angle $= 180^\circ - \text{Measure of each exterior angle}$

Measure of each interior angle $= 180^\circ - 45^\circ$

Measure of each interior angle $= 135^\circ$

This confirms the result obtained earlier.

Given:

A regular pentagon.

A regular decagon.

To Find:

1. The measure of an exterior angle of the regular pentagon.

2. The measure of an exterior angle of the regular decagon.

3. The ratio between these two angles.

Solution:

The measure of each exterior angle of a regular polygon with $n$ sides is given by the formula:

Measure of each exterior angle $= \frac{360^\circ}{n}$

For the regular pentagon:

A pentagon has $n = 5$ sides.

Measure of each exterior angle of the pentagon $= \frac{360^\circ}{5}$

Measure of each exterior angle of the pentagon $= 72^\circ$

For the regular decagon:

A decagon has $n = 10$ sides.

Measure of each exterior angle of the decagon $= \frac{360^\circ}{10}$

Measure of each exterior angle of the decagon $= 36^\circ$

Ratio between the two angles:

Ratio $= \frac{\text{Exterior angle of pentagon}}{\text{Exterior angle of decagon}}$

Ratio $= \frac{72^\circ}{36^\circ}$

Ratio $= \frac{72}{36}$

Ratio $= 2$

The ratio can also be expressed as $2:1$.

The measure of an exterior angle of a regular pentagon is $72^\circ$.

The measure of an exterior angle of a regular decagon is $36^\circ$.

The ratio between these two angles is $2:1$ or $2$.

Given:

From the figure, we have a quadrilateral with interior angles $\angle A = 70^\circ$, $\angle B = 120^\circ$, and $\angle D = 110^\circ$.

The exterior angle at vertex C is $x$. Let the interior angle at C be $\angle C_{\text{int}}$.

To Find:

The value of x.

Solution:

The sum of the interior angles of any quadrilateral is $360^\circ$.

So, $\angle A + \angle B + \angle C_{\text{int}} + \angle D = 360^\circ$

Substitute the given values of the angles:

$70^\circ + 120^\circ + \angle C_{\text{int}} + 110^\circ = 360^\circ$

Add the known angles:

$(70^\circ + 120^\circ + 110^\circ) + \angle C_{\text{int}} = 360^\circ$

$300^\circ + \angle C_{\text{int}} = 360^\circ$

Subtract $300^\circ$ from both sides to find the interior angle at C:

$\angle C_{\text{int}} = 360^\circ - 300^\circ$

$\angle C_{\text{int}} = 60^\circ$

The interior angle at vertex C ($\angle C_{\text{int}}$) and the exterior angle at vertex C (x) form a linear pair.

The sum of angles in a linear pair is $180^\circ$.

So, $\angle C_{\text{int}} + x = 180^\circ$

Substitute the value of $\angle C_{\text{int}}$:

$60^\circ + x = 180^\circ$

Subtract $60^\circ$ from both sides:

$x = 180^\circ - 60^\circ$

$x = 120^\circ$

The value of x is $120^\circ$.

Given:

A quadrilateral has three equal angles.

The fourth angle has a measure of $120^\circ$.

To Find:

The measure of each of the three equal angles.

Solution:

Let the measure of each of the three equal angles be $x^\circ$.

The angles of the quadrilateral are $x^\circ, x^\circ, x^\circ,$ and $120^\circ$.

The sum of the interior angles of any quadrilateral is $360^\circ$.

Therefore, we can write the equation:

$x + x + x + 120^\circ = 360^\circ$

Combine the terms with x:

$3x + 120^\circ = 360^\circ$

Subtract $120^\circ$ from both sides of the equation:

$3x = 360^\circ - 120^\circ$

$3x = 240^\circ$

Divide both sides by 3:

$x = \frac{240^\circ}{3}$

$x = 80^\circ$

Thus, the measure of each of the three equal angles is $80^\circ$.

The measures of the angles of the quadrilateral are $80^\circ, 80^\circ, 80^\circ,$ and $120^\circ$.

Sum of angles $= 80^\circ + 80^\circ + 80^\circ + 120^\circ = 240^\circ + 120^\circ = 360^\circ$.

Given:

Quadrilateral HOPE.

PS is the bisector of $\angle$P.

ES is the bisector of $\angle$E.

PS and ES meet at point S.

To Give Reason:

To state a property or relationship related to the intersection point S of the bisectors of the opposite angles $\angle$P and $\angle$E.

Reason:

The point S, where the bisectors of two opposite angles of a quadrilateral meet, has a specific property related to its distance from the sides of the quadrilateral.

By the property of an angle bisector, any point on the angle bisector is equidistant from the two sides forming the angle.

Since S lies on the bisector of $\angle$P (which is formed by sides HP and PO), point S is equidistant from the sides HP and PO.

Since S lies on the bisector of $\angle$E (which is formed by sides HE and EO), point S is equidistant from the sides HE and EO.

Thus, the reason is based on the fundamental property of angle bisectors: a point on an angle bisector is equidistant from the arms of the angle.

Additionally, there is a theorem relating the angle formed by the bisectors of two opposite angles of a quadrilateral to the other two angles.

The angle between the bisectors of two opposite angles of a quadrilateral is half the absolute difference of the other two angles.

In quadrilateral HOPE, the bisectors of opposite angles $\angle$P and $\angle$E meet at S. The other two angles are $\angle$H and $\angle$O.

The angle formed by the bisectors is $\angle$PSE (or its vertically opposite angle).

Therefore, $\angle PSE = \frac{1}{2} |\angle H - \angle O|$.

This theorem provides a significant relationship regarding the angle at the intersection point S based on the measures of the other two angles of the quadrilateral.

Given:

ABCD is a parallelogram.

From the figure, the given angles are:

$\angle BAC = 30^\circ$

$\angle AOB = 110^\circ$ (where O is the intersection of diagonals AC and BD)

$\angle BCD = x$

$\angle CAD = x$

$\angle ACD = y$

$\angle ACB = z$

To Find:

The values of x, y and z.

Solution:

In a parallelogram, opposite sides are parallel.

Since AB || DC, consider the diagonal AC as a transversal.

The alternate interior angles are equal.

Given $\angle BAC = 30^\circ$ and $\angle ACD = y$.

So, $y = 30^\circ$.

Since AD || BC, consider the diagonal AC as a transversal.

The alternate interior angles are equal.

Given $\angle CAD = x$ and $\angle ACB = z$.

So, $z = x$.

Consider the vertex C. The interior angle $\angle BCD$ is the sum of angles $\angle ACD$ and $\angle ACB$.

$\angle BCD = \angle ACD + \angle ACB$

Substitute the variables from the figure: $\angle BCD = x$, $\angle ACD = y$, $\angle ACB = z$.

So, $x = y + z$.

Now substitute the values we found for y and z into this equation:

Substitute $y = 30^\circ$:

$x = 30^\circ + z$

Substitute $z = x$:

$x = 30^\circ + x$

Subtract x from both sides:

$x - x = 30^\circ$

$0 = 30^\circ$

This is a contradiction. The given information in the problem statement and figure is inconsistent, making it impossible to find values for x, y, and z that satisfy all conditions simultaneously in a parallelogram.

Analysis of the Contradiction:

The contradiction arises directly from the angle labels in the figure ($\angle CAD = x$, $\angle BCD = x$, $\angle ACD = y$, $\angle ACB = z$) combined with the parallelogram properties ($\angle ACD = \angle BAC = 30^\circ \implies y=30^\circ$ and $\angle ACB = \angle CAD \implies z=x$) and the angle addition rule ($\angle BCD = \angle ACD + \angle ACB \implies x = y+z$). Substituting the values $y=30^\circ$ and $z=x$ into $x=y+z$ leads to $x = 30^\circ + x$, which is $0 = 30^\circ$.

This indicates that the problem as stated has an error.

Likely Intended Problem (Assuming a Typo):

Given the common types of problems and the presence of angle $30^\circ$ and $110^\circ$ related to the diagonals, a plausible intended scenario leads to simple integer angle values.

Let's analyze the angles formed by the diagonals if we ignore the explicit condition $\angle BCD = x$ as leading to a contradiction, and assume we just need to find $x, y, z$ based on other consistent conditions.

From $\triangle AOB$, $\angle OAB = 30^\circ$, $\angle AOB = 110^\circ$. $\angle ABO = 180^\circ - 110^\circ - 30^\circ = 40^\circ$.

From $\triangle AOD$, $\angle OAD = x$, $\angle DOA = 180^\circ - 110^\circ = 70^\circ$. $\angle ODA = 180^\circ - 70^\circ - x = 110^\circ - x$.

From $\triangle BOC$, $\angle OCB = z$, $\angle BOC = 70^\circ$. $\angle OBC = 180^\circ - 70^\circ - z = 110^\circ - z$.

From $\triangle COD$, $\angle OCD = y = 30^\circ$, $\angle COD = 110^\circ$. $\angle ODC = 180^\circ - 110^\circ - 30^\circ = 40^\circ$.

Parallelogram property: $\angle B = \angle D$. $\angle B = \angle ABO + \angle OBC = 40^\circ + (110^\circ - z)$. $\angle D = \angle ODA + \angle ODC = (110^\circ - x) + 40^\circ$.

$40^\circ + 110^\circ - z = 110^\circ - x + 40^\circ \implies 150^\circ - z = 150^\circ - x \implies z = x$. This was already found from alternate interior angles.

Consider the possibility that the angles $\angle CAD$, $\angle ACB$, and $\angle ACD$ might have simple relationships, or that a specific type of parallelogram (like one formed by equilateral triangles or isosceles triangles) was intended.

If we assume the consistent case where $x=30^\circ, y=30^\circ, z=30^\circ$ (as explored in thought process), then $\angle CAD=30^\circ, \angle ACD=30^\circ, \angle ACB=30^\circ$. In this case, $\angle A = 30+30=60^\circ$, $\angle C=30+30=60^\circ$, $\angle B=\angle D=120^\circ$. The angles in $\triangle AOB$ would be $\angle OAB=30$, $\angle ABO=40$, $\angle AOB=110$. This is consistent ($30+40+110=180$).

If $(x, y, z) = (30^\circ, 30^\circ, 30^\circ)$, then $\angle CAD=30^\circ$, $\angle ACD=30^\circ$, $\angle ACB=30^\circ$. The label $\angle BCD = x$ would then mean $\angle BCD = 30^\circ$. But $\angle BCD = \angle ACD + \angle ACB = 30^\circ + 30^\circ = 60^\circ$. This again contradicts the label $\angle BCD = x = 30^\circ$.

The only scenario where the values $(x, y, z) = (30^\circ, 30^\circ, 30^\circ)$ make sense is if the label $\angle CAD = x$ implies $x=30^\circ$, the label $\angle ACD = y$ implies $y=30^\circ$, and the label $\angle ACB = z$ implies $z=30^\circ$. And the label $\angle BCD = x$ is simply a label for the entire angle, whose value is calculated as $y+z = 30+30=60^\circ$, not necessarily equal to the variable $x=30^\circ$. However, this contradicts the question "Find the value of x, y and z" if 'x' is meant to be a single value throughout.

Given the clear contradiction $0=30^\circ$ resulting from the direct interpretation, and the expectation of a unique numerical solution, there is a fundamental error in the problem statement or the figure labels.

Based on the high probability of a typo and the consistency of angles (30, 40, 70) in the diagrammatic representation of triangle angles around O, the most likely intended values derived from a consistent problem are:

y = $30^\circ$ (from $\angle ACD = \angle BAC$)

z = $30^\circ$ (from $\angle ACB = \angle CAD$ and assuming $x=30^\circ$ was intended for $\angle CAD$)

x = $30^\circ$ (assuming $\angle CAD = 30^\circ$ was intended)

With these values, $\angle BCD = y+z = 30+30=60^\circ$. If the label $\angle BCD=x$ was intended to mean $\angle BCD = \angle CAD$, this leads back to the contradiction $60^\circ = 30^\circ$.

Assuming the values that create a consistent parallelogram geometry from the given numerical angles are expected, and acknowledging the contradiction caused by the variable 'x' appearing in two inconsistent places:

Based on $\angle BAC = 30^\circ$, $\angle AOB = 110^\circ$, and parallelogram properties, the angles are consistent with:

$\angle CAD = 30^\circ \implies x = 30^\circ$

$\angle ACD = 30^\circ \implies y = 30^\circ$

$\angle ACB = 30^\circ \implies z = 30^\circ$

This interpretation resolves the triangle angle sums and alternate interior angles consistently ($30,40,70,80,110$). It implies that the variable label 'x' on $\angle BCD$ (which would be $60^\circ$ in this case) is a typo and should not be equated to the variable 'x' on $\angle CAD$ (which is $30^\circ$).

Given the contradictory nature, the problem cannot be solved as strictly stated. However, if a numerical answer is required, the values $x=30^\circ, y=30^\circ, z=30^\circ$ fit a consistent parallelogram structure related to the given numerical angles.

Value of x = $30^\circ$

Value of y = $30^\circ$

Value of z = $30^\circ$

Given:

A quadrilateral whose diagonals are perpendicular to each other.

To Answer:

Is such a quadrilateral always a rhombus? Provide a figure for justification.

Solution:

No, a quadrilateral whose diagonals are perpendicular to each other is not always a rhombus.

Justification:

A rhombus is a quadrilateral in which all four sides are equal in length.

One of the properties of a rhombus is that its diagonals are perpendicular bisectors of each other.

If the diagonals of a quadrilateral are perpendicular, it does not necessarily mean that they bisect each other, nor does it mean that all four sides are equal.

Consider a kite. A kite is a quadrilateral with two distinct pairs of equal-length adjacent sides. The diagonals of a kite are always perpendicular to each other, and one of the diagonals is the perpendicular bisector of the other.

In a kite, the diagonals are perpendicular, but typically, the diagonals do not bisect each other (only one does), and the four sides are not equal (only adjacent pairs are equal). Since a rhombus requires all four sides to be equal, a kite with perpendicular diagonals is not a rhombus unless it is also a rhombus (in which case it's a square).

Figure (Illustration of a Kite):

Let ABCD be a quadrilateral.

Let the diagonals AC and BD intersect at point O.

Assume AC $\perp$ BD, so $\angle AOB = \angle BOC = \angle COD = \angle DOA = 90^\circ$.

Consider the case where ABCD is a kite, with AB = AD and CB = CD. The diagonal AC is the axis of symmetry and it perpendicularly bisects BD. So, BO = OD, and AC $\perp$ BD.

However, in a general kite, AB $\neq$ BC (unless the kite is a rhombus/square). Thus, the sides are not all equal.

For example, a quadrilateral with vertices at A(0, 3), B(-2, 0), C(0, -3), and D(2, 0) is a kite. The diagonals are AC (on the y-axis) and BD (on the x-axis). They are perpendicular at the origin (0, 0). The length of AC is 6, and the length of BD is 4. They are perpendicular.

Let's calculate side lengths:

$AB = \sqrt{(-2-0)^2 + (0-3)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}$

$AD = \sqrt{(2-0)^2 + (0-3)^2} = \sqrt{(2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}$

$CB = \sqrt{(-2-0)^2 + (0-(-3))^2} = \sqrt{(-2)^2 + (3)^2} = \sqrt{4 + 9} = \sqrt{13}$

$CD = \sqrt{(2-0)^2 + (0-(-3))^2} = \sqrt{(2)^2 + (3)^2} = \sqrt{4 + 9} = \sqrt{13}$

In this specific example, it turns out AB = AD = CB = CD = $\sqrt{13}$. This particular kite is also a rhombus (and since the diagonals are perpendicular bisectors of each other, it's even a square). My example of a kite was poor.

Let's use a better kite example where sides are not equal.

Vertices at A(0, 4), B(-3, 0), C(0, -2), D(3, 0).

Diagonals are AC (on y-axis) and BD (on x-axis). They are perpendicular at the origin O(0,0).

Side lengths:

$AB = \sqrt{(-3-0)^2 + (0-4)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

$AD = \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

$CB = \sqrt{(-3-0)^2 + (0-(-2))^2} = \sqrt{(-3)^2 + (2)^2} = \sqrt{9 + 4} = \sqrt{13}$

$CD = \sqrt{(3-0)^2 + (0-(-2))^2} = \sqrt{(3)^2 + (2)^2} = \sqrt{9 + 4} = \sqrt{13}$

Here, AB = AD = 5 and CB = CD = $\sqrt{13}$. The diagonals are perpendicular, but the sides are not all equal (5 $\neq \sqrt{13}$). This is a kite that is not a rhombus.

Therefore, a quadrilateral with perpendicular diagonals is not always a rhombus.

Given:

ABCD is a trapezium.

AB || CD.

The ratio of angles $\angle A$ and $\angle D$ is $\angle A : \angle D = 2 : 1$.

The ratio of angles $\angle B$ and $\angle C$ is $\angle B : \angle C = 7 : 5$.

To Find:

The measure of each angle of the trapezium: $\angle A, \angle B, \angle C, \angle D$.

Solution:

In a trapezium, consecutive angles between the parallel sides (on the same leg) are supplementary.

Since AB || CD, the leg AD forms a transversal. Thus, $\angle A$ and $\angle D$ are supplementary.

$\angle A + \angle D = 180^\circ$

The ratio $\angle A : \angle D = 2 : 1$. Let $\angle A = 2k$ and $\angle D = 1k = k$ for some constant $k$.

Substitute these into the supplementary equation:

$2k + k = 180^\circ$

$3k = 180^\circ$

$k = \frac{180^\circ}{3}$

$k = 60^\circ$

Now find the measures of $\angle A$ and $\angle D$:

$\angle A = 2k = 2 \times 60^\circ = 120^\circ$

$\angle D = k = 60^\circ$

Similarly, for the leg BC, $\angle B$ and $\angle C$ are supplementary.

$\angle B + \angle C = 180^\circ$

The ratio $\angle B : \angle C = 7 : 5$. Let $\angle B = 7m$ and $\angle C = 5m$ for some constant $m$.

Substitute these into the supplementary equation:

$7m + 5m = 180^\circ$

$12m = 180^\circ$

$m = \frac{180^\circ}{12}$

$m = 15^\circ$

Now find the measures of $\angle B$ and $\angle C$:

$\angle B = 7m = 7 \times 15^\circ = 105^\circ$

$\angle C = 5m = 5 \times 15^\circ = 75^\circ$

The angles of the trapezium are $\angle A = 120^\circ$, $\angle B = 105^\circ$, $\angle C = 75^\circ$, and $\angle D = 60^\circ$.

Given:

Line $l$ is parallel to line $m$ ($l || m$).

Transversal $p$ intersects $l$ at X and $m$ at Y.

Let the interior angles at X be $\angle AXY$ and $\angle BXY$, where A and B are points on line $l$ such that A-X-B.

Let the interior angles at Y be $\angle CYX$ and $\angle DYX$, where C and D are points on line $m$ such that C-Y-D.

P is the intersection of the bisector of $\angle AXY$ and the bisector of $\angle CYX$.

Q is the intersection of the bisector of $\angle BXY$ and the bisector of $\angle DYX$.

To Determine:

Is quadrilateral PXQY a rectangle? Give reason.

Proof:

Since $l || m$ and $p$ is a transversal:

XP is the bisector of $\angle AXY$. Therefore,

$\angle PXY = \frac{1}{2} \angle AXY$

YP is the bisector of $\angle CYX$. Therefore,

$\angle PYX = \frac{1}{2} \angle CYX$

Consider $\triangle$PXY. The sum of angles in a triangle is $180^\circ$.

Substitute the half-angle values:

$\angle$XPY + $\frac{1}{2} \angle AXY + \frac{1}{2} \angle CYX = $180^\circ$

$\angle$XPY + $\frac{1}{2} (\angle AXY + \angle CYX) = $180^\circ$

Substitute $\angle AXY + \angle CYX = 180^\circ$:

$\angle$XPY + $\frac{1}{2} (180^\circ) = $180^\circ$

$\angle$XPY + $90^\circ = $180^\circ$

$\angle$XPY = $180^\circ - 90^\circ$

$\angle$XPY = $90^\circ$

Thus, the angle at vertex P in quadrilateral PXQY is $90^\circ$.

Similarly, consider the consecutive interior angles on the other side of the transversal:

XQ is the bisector of $\angle BXY$. Therefore,

$\angle QXY = \frac{1}{2} \angle BXY$

YQ is the bisector of $\angle DYX$. Therefore,

$\angle QYX = \frac{1}{2} \angle DYX$

Consider $\triangle$QXY. The sum of angles in a triangle is $180^\circ$.

Substitute the half-angle values:

$\angle$XQY + $\frac{1}{2} \angle BXY + \frac{1}{2} \angle DYX = $180^\circ$

$\angle$XQY + $\frac{1}{2} (\angle BXY + \angle DYX) = $180^\circ$

Substitute $\angle BXY + \angle DYX = 180^\circ$:

$\angle$XQY + $\frac{1}{2} (180^\circ) = $180^\circ$

$\angle$XQY + $90^\circ = $180^\circ$

$\angle$XQY = $180^\circ - 90^\circ$

$\angle$XQY = $90^\circ$

Thus, the angle at vertex Q in quadrilateral PXQY is $90^\circ$.

Now consider the angles at vertex X in quadrilateral PXQY. This is $\angle$PXQ.

Angles $\angle AXY$ and $\angle BXY$ are adjacent angles on the straight line $l$ at point X.

XP is the bisector of $\angle AXY$, and XQ is the bisector of $\angle BXY$. The angle formed by the bisectors of two adjacent angles forming a linear pair is $90^\circ$.

Specifically, $\angle PXQ = \angle PXY + \angle QXY$ (since ray XY is between rays XP and XQ).

$\angle PXY = \frac{1}{2} \angle AXY$ and $\angle QXY = \frac{1}{2} \angle BXY$.

$\angle PXQ = \frac{1}{2} \angle AXY + \frac{1}{2} \angle BXY$

$\angle PXQ = \frac{1}{2} (\angle AXY + \angle BXY)$

Substitute $\angle AXY + \angle BXY = 180^\circ$:

$\angle$PXQ = $\frac{1}{2} (180^\circ)$

$\angle$PXQ = $90^\circ$

Thus, the angle at vertex X in quadrilateral PXQY is $90^\circ$.

Similarly, consider the angles at vertex Y in quadrilateral PXQY. This is $\angle$PYQ.

Angles $\angle CYX$ and $\angle DYX$ are adjacent angles on the straight line $m$ at point Y.

YP is the bisector of $\angle CYX$, and YQ is the bisector of $\angle DYX$. The angle formed by the bisectors of two adjacent angles forming a linear pair is $90^\circ$.

Specifically, $\angle PYQ = \angle PYX + \angle QYX$ (since ray YX is between rays YP and YQ).

$\angle PYX = \frac{1}{2} \angle CYX$ and $\angle QYX = \frac{1}{2} \angle DYX$.

$\angle PYQ = \frac{1}{2} \angle CYX + \frac{1}{2} \angle DYX$

$\angle PYQ = \frac{1}{2} (\angle CYX + \angle DYX)$

Substitute $\angle CYX + \angle DYX = 180^\circ$:

$\angle$PYQ = $\frac{1}{2} (180^\circ)$

$\angle$PYQ = $90^\circ$

Thus, the angle at vertex Y in quadrilateral PXQY is $90^\circ$.

The angles of the quadrilateral PXQY are $\angle$XPY = $90^\circ$, $\angle$XQY = $90^\circ$, $\angle$PXQ = $90^\circ$, and $\angle$PYQ = $90^\circ$.

Since all four interior angles of quadrilateral PXQY are $90^\circ$, it is a rectangle.

Reason:

A quadrilateral is a rectangle if and only if all its interior angles are $90^\circ$. We have shown that each interior angle of the quadrilateral PXQY measures $90^\circ$.

Given:

ABCD is a parallelogram.

AX is the bisector of $\angle A$, intersecting CD at X.

CY is the bisector of $\angle C$, intersecting AB at Y.

To Determine:

Is quadrilateral AXCY a parallelogram? Give reason.

Proof:

In parallelogram ABCD, we know that opposite sides are parallel and equal, and opposite angles are equal.

So, AB || CD and AB = CD.

Also, AD || BC and AD = BC.

And, $\angle A = \angle C$ and $\angle B = \angle D$.

Since AX is the bisector of $\angle A$, we have $\angle DAX = \angle XAB = \frac{1}{2} \angle A$.

Since CY is the bisector of $\angle C$, we have $\angle BCY = \angle YCD = \frac{1}{2} \angle C$.

As $\angle A = \angle C$, it follows that $\frac{1}{2} \angle A = \frac{1}{2} \angle C$, so $\angle XAB = \angle YCD$.

Consider the parallel lines AB and CD cut by the transversal AX.

The alternate interior angles are equal.

Thus, $\angle XAB = \angle AXD$ (Alternate interior angles, since AB || CD).

Since $\angle XAB = \angle DAX$, we have $\angle DAX = \angle AXD$.

In $\triangle ADX$, the angles opposite to sides AD and DX are $\angle AXD$ and $\angle DAX$ respectively.

Since $\angle DAX = \angle AXD$, the triangle $\triangle ADX$ is isosceles.

Therefore, the sides opposite these angles are equal: AD = DX.

Consider the parallel lines AB and CD cut by the transversal CY.

The alternate interior angles are equal.

Thus, $\angle YCD = \angle CYB$ (Alternate interior angles, since CD || AB).

Since $\angle YCD = \angle BCY$, we have $\angle BCY = \angle CYB$.

In $\triangle BCY$, the angles opposite to sides BC and BY are $\angle CYB$ and $\angle BCY$ respectively.

Since $\angle BCY = \angle CYB$, the triangle $\triangle BCY$ is isosceles.

Therefore, the sides opposite these angles are equal: BC = BY.

We know that ABCD is a parallelogram, so AD = BC (opposite sides are equal).

Since AD = DX and BC = BY, and AD = BC, we can conclude that DX = BY.

Now consider the sides AB and CD of the parallelogram ABCD.

AB = CD (opposite sides of parallelogram).

We can write AB as the sum of segments AY and YB: AB = AY + YB.

We can write CD as the sum of segments CX and XD: CD = CX + XD.

So, AY + YB = CX + XD.

Substitute YB = BY and XD = DX: AY + BY = CX + DX.

Since BY = DX, we can substitute DX for BY:

AY + DX = CX + DX

Subtract DX from both sides of the equation:

AY = CX.

Now consider the quadrilateral AXCY.

We have shown that AY = CX.

Also, since AB || CD and Y is on AB and X is on CD, the segment AY is parallel to the segment CX (they lie on parallel lines).

So, in quadrilateral AXCY, one pair of opposite sides (AY and CX) is both equal and parallel.

Reason:

A quadrilateral is a parallelogram if one pair of opposite sides is both parallel and equal in length.

Since AY || CX and AY = CX, quadrilateral AXCY is a parallelogram.

Given:

ABCD is a parallelogram.

Let the diagonal be AC.

Diagonal AC bisects $\angle A$, meaning $\angle DAC = \angle BAC$.

To Determine:

Does the diagonal AC also bisect $\angle C$?

Proof:

In a parallelogram ABCD, we know that opposite sides are parallel.

So, AB || CD and AD || BC.

Consider the transversal AC intersecting the parallel lines AB and CD.

The alternate interior angles are equal.

Now, consider the transversal AC intersecting the parallel lines AD and BC.

The alternate interior angles are equal.

We are given that the diagonal AC bisects $\angle A$.

Now, let's compare the angles that make up $\angle C$ ($\angle ACD$ and $\angle BCA$).

We have $\angle ACD = \angle BAC$ and $\angle BCA = \angle DAC$.

Since $\angle DAC = \angle BAC$, we can substitute this into the equation for $\angle BCA$:

$\angle BCA = \angle BAC$ (since $\angle DAC = \angle BAC$)

So we have $\angle ACD = \angle BAC$ and $\angle BCA = \angle BAC$.

This means $\angle ACD = \angle BCA$.

Since the diagonal AC divides $\angle C$ into two equal angles ($\angle ACD$ and $\angle BCA$), it bisects $\angle C$.

Reason:

Yes, the diagonal will also bisect the other angle. This is because when a diagonal bisects one angle of a parallelogram, it makes the two alternate interior angles (formed by the diagonal with the parallel sides) equal. These alternate interior angles are precisely the two parts of the opposite angle. Since these two parts are equal, the diagonal must bisect the opposite angle as well.

Also, if a diagonal of a parallelogram bisects an angle, it implies that the adjacent sides meeting at that vertex are equal (e.g., if $\angle DAC = \angle BAC$, then $\triangle ADC$ and $\triangle ABC$ are isosceles, leading to AD=CD and AB=BC). Since opposite sides are already equal (AD=BC, AB=CD), this means all four sides are equal (AD=BC=CD=AB). A parallelogram with all four sides equal is a rhombus. In a rhombus, the diagonals always bisect the angles.

Given:

A parallelogram.

The angle between the two altitudes drawn from the vertex of an obtuse angle is $45^\circ$.

To Find:

The measure of each angle of the parallelogram.

Solution:

Let the parallelogram be ABCD, and let $\angle B$ be the obtuse angle.

Let BE and BF be the altitudes from vertex B to the lines containing the adjacent sides AD and CD, respectively.

Since $\angle B$ is obtuse, the altitudes from B will fall on the extensions of the sides AD and CD.

So, BE $\perp$ line AD (meaning BE $\perp$ AD extended) and BF $\perp$ line CD (meaning BF $\perp$ CD extended).

Therefore, $\angle BEA = 90^\circ$ and $\angle BFC = 90^\circ$. If we consider the points where the altitudes meet the lines containing the sides, let's denote them E and F such that E is on the line containing AD and F is on the line containing CD.

Given that the angle between the altitudes is $45^\circ$, we have $\angle EBF = 45^\circ$.

Consider the quadrilateral formed by the vertex B, the feet of the altitudes E and F, and the vertex D (which is opposite to B). The vertices of this quadrilateral are B, E, D, and F.

In quadrilateral BEDF, the angles are $\angle EBF$, $\angle BED$, $\angle BFD$, and $\angle EDF$.

We have $\angle BED = 90^\circ$ and $\angle BFD = 90^\circ$.

The angle $\angle EDF$ is formed by the lines AD and CD, which are sides of the parallelogram. $\angle EDF$ is the same as the interior angle $\angle ADC$ of the parallelogram, i.e., $\angle EDF = \angle D$.

The sum of the interior angles of a quadrilateral is $360^\circ$.

So, in quadrilateral BEDF:

$\angle EBF + \angle BED + \angle EDF + \angle BFD = 360^\circ$

Substitute the known values:

$45^\circ + 90^\circ + \angle D + 90^\circ = 360^\circ$

Combine the constant terms:

$225^\circ + \angle D = 360^\circ$

Subtract $225^\circ$ from both sides to find $\angle D$:

$\angle D = 360^\circ - 225^\circ$

$\angle D = 135^\circ$

In a parallelogram, opposite angles are equal.

So, $\angle B = \angle D = 135^\circ$. This confirms $\angle B$ is obtuse as assumed.

Consecutive angles in a parallelogram are supplementary (sum up to $180^\circ$).

So, $\angle A + \angle B = 180^\circ$

$\angle A + 135^\circ = 180^\circ$

$\angle A = 180^\circ - 135^\circ$

$\angle A = 45^\circ$

Since opposite angles are equal, $\angle C = \angle A = 45^\circ$.

The angles of the parallelogram are $45^\circ, 135^\circ, 45^\circ, 135^\circ$.

Given:

ABCD is a rhombus.

The perpendicular bisector of side AB passes through vertex D.

To Find:

The angles of the rhombus ($\angle A, \angle B, \angle C, \angle D$).

Solution:

We are given that the perpendicular bisector of side AB passes through point D.

A fundamental property of a perpendicular bisector is that any point on it is equidistant from the endpoints of the segment it bisects.

Since D lies on the perpendicular bisector of AB, the distance from D to A is equal to the distance from D to B.

Thus, DA = DB.

ABCD is a rhombus, which is a type of parallelogram with all four sides equal in length.

So, AB = BC = CD = DA.

From the property of the rhombus, we have DA = AB.

Combining the two equalities (DA = DB and DA = AB), we get:

AB = DB = DA

Consider the triangle $\triangle$ABD. The sides of this triangle are AB, BD, and DA.

Since AB = BD = DA, $\triangle$ABD is an equilateral triangle.

In an equilateral triangle, all interior angles are equal, and each measures $60^\circ$.

Therefore, the angles in $\triangle$ABD are:

$\angle DAB = 60^\circ$

$\angle ABD = 60^\circ$

$\angle BDA = 60^\circ$

The angle $\angle DAB$ is one of the interior angles of the rhombus ABCD. So, $\angle A = 60^\circ$.

In a parallelogram (and thus in a rhombus), opposite angles are equal.

So, $\angle C = \angle A = 60^\circ$.

Also, consecutive angles in a parallelogram are supplementary (sum up to $180^\circ$).

So, $\angle A + \angle B = 180^\circ$.

$60^\circ + \angle B = 180^\circ$

$\angle B = 180^\circ - 60^\circ$

$\angle B = 120^\circ$

Since opposite angles are equal, $\angle D = \angle B = 120^\circ$.

Thus, the angles of the rhombus are $60^\circ, 120^\circ, 60^\circ,$ and $120^\circ$.

The angles of the rhombus are $\angle A = 60^\circ$, $\angle B = 120^\circ$, $\angle C = 60^\circ$, and $\angle D = 120^\circ$.

Given:

ABCD is a parallelogram.

Point P is on side AB.

Point Q is on side AD.

PRQA is a parallelogram formed by points P, R, Q, and A.

The measure of $\angle C$ in parallelogram ABCD is $45^\circ$.

To Find:

The measure of $\angle R$ in parallelogram PRQA.

Solution:

In parallelogram ABCD, opposite angles are equal.

Therefore, $\angle A$ (in parallelogram ABCD) = $\angle C$ (in parallelogram ABCD).

Given that $\angle C = 45^\circ$, we have:

$\angle A = 45^\circ$

The angle $\angle A$ in parallelogram ABCD is the angle $\angle BAD$.

Consider the parallelogram PRQA.

The problem states that points P and Q are taken on sides AB and AD respectively, and parallelogram PRQA is formed.

Since P is on AB and Q is on AD, and A is a vertex of parallelogram PRQA, this implies that the adjacent sides of parallelogram PRQA originating from A are AP and AQ.

So, parallelogram PRQA has vertices A, P, R, Q (or A, Q, R, P).

The angle at vertex A in parallelogram PRQA is $\angle PAQ$.

Since P is on side AB and Q is on side AD, the angle $\angle PAQ$ is the same as the angle $\angle BAD$ of the parallelogram ABCD.

So, $\angle A$ (in parallelogram PRQA) = $\angle BAD$ (in parallelogram ABCD).

We found that $\angle BAD = 45^\circ$.

Thus, $\angle A$ (in parallelogram PRQA) = $45^\circ$.

In parallelogram PRQA, opposite angles are equal.

The angle $\angle R$ in parallelogram PRQA is opposite to the angle $\angle A$ (or $\angle PAQ$).

Therefore, $\angle R$ (in parallelogram PRQA) = $\angle A$ (in parallelogram PRQA).

$\angle R = 45^\circ$.

The measure of $\angle R$ is $45^\circ$.

Given:

ABCD is a parallelogram.

The angle bisector of $\angle A$ bisects side BC.

To Determine:

Will the angle bisector of $\angle B$ also bisect side AD? Give reason.

Proof:

Let AM be the angle bisector of $\angle A$, where M is a point on BC such that AM bisects BC.

Since AM bisects BC, M is the midpoint of BC.

In parallelogram ABCD, AD || BC and AB || DC.

Consider the parallel lines AB and DC cut by the transversal AM.

$\angle BAM = \angle DMA$ (Alternate interior angles, AB || DC).

Since AM is the angle bisector of $\angle A$, $\angle BAM = \angle DAM$.

From the above two equalities, we get $\angle DAM = \angle DMA$.

In $\triangle ADM$, the angles opposite to sides AD and AM are $\angle DMA$ and $\angle DAM$ respectively. Actually, angles opposite to sides AM and AD are $\angle ADM$ and $\angle AMD$ respectively. The angles opposite to sides AD and DM are $\angle AMD$ and $\angle DAM$ respectively. The angles opposite to sides AM and DM are $\angle ADM$ and $\angle DAM$ respectively.

Let's reconsider the angles in $\triangle ADM$. The angles are $\angle DAM$, $\angle ADM$ (which is $\angle D$ of the parallelogram), and $\angle AMD$.

We have $\angle DAM = \angle DMA$. This means $\triangle ADM$ is an isosceles triangle with sides opposite to these angles being equal.

The side opposite $\angle DMA$ is AD.

The side opposite $\angle DAM$ is DM.

So, AD = DM.

This contradicts the fact that M is on BC. Let's re-examine the alternate interior angles.

Consider transversal AM cutting parallel lines AB and CD. This creates $\angle BAM$ and $\angle DMA$. This is not correct as AM is the transversal and AB || CD. The alternate interior angles should be related to the transversal cutting the parallel lines. AM cuts parallel lines AB and CD. The alternate interior angles formed are $\angle BAM$ and $\angle AXD$ if X is on CD and AX is the bisector. Here the bisector AM goes to M on BC.

Let's use AD || BC and transversal AM.

$\angle DAM = \angle AMB$ (Alternate interior angles, AD || BC).

Given that AM bisects $\angle A$, $\angle DAM = \angle BAM$.

From the above two equalities, $\angle AMB = \angle BAM$.

In $\triangle ABM$, we have $\angle AMB = \angle BAM$.

Therefore, $\triangle ABM$ is an isosceles triangle with sides opposite to these equal angles being equal.

The side opposite $\angle AMB$ is AB.

The side opposite $\angle BAM$ is BM.

So, AB = BM.

Since M is the midpoint of BC, we have BC = 2 * BM.

Substituting BM = AB, we get BC = 2 * AB.

In a parallelogram, opposite sides are equal, so AD = BC.

Therefore, AD = 2 * AB.

Now consider the angle bisector of $\angle B$. Let BN be the angle bisector of $\angle B$, where N is a point on AD.

In parallelogram ABCD, AD || BC and AB || DC.

Consider the parallel lines AD and BC cut by the transversal BN.

$\angle CBN = \angle BNA$ (Alternate interior angles, AD || BC).

Since BN is the angle bisector of $\angle B$, $\angle CBN = \angle ABN$.

From the above two equalities, $\angle BNA = \angle ABN$.

In $\triangle ABN$, we have $\angle BNA = \angle ABN$.

Therefore, $\triangle ABN$ is an isosceles triangle with sides opposite to these equal angles being equal.

The side opposite $\angle BNA$ is AB.

The side opposite $\angle ABN$ is AN.

So, AB = AN.

We previously deduced that AD = 2 * AB.

Substitute AB = AN into this equation:

AD = 2 * AN.

This implies that AN = $\frac{1}{2}$ AD.

Since N is on the side AD and AN is half the length of AD, N must be the midpoint of AD.

Reason:

Yes, the angle bisector of $\angle B$ will also bisect AD.

Reason: When the angle bisector of $\angle A$ bisects side BC at M, it implies that $\triangle ABM$ is isosceles with AB = BM. Since M is the midpoint of BC, BC = 2 * BM = 2 * AB. As AD = BC in a parallelogram, we have AD = 2 * AB.

Now, consider the angle bisector of $\angle B$, say BN, intersecting AD at N. Using alternate interior angles (AD || BC, transversal BN), $\angle CBN = \angle BNA$. Since BN bisects $\angle B$, $\angle CBN = \angle ABN$. Thus, $\angle ABN = \angle BNA$. This makes $\triangle ABN$ isosceles with AB = AN. Since we know AD = 2 * AB, substituting AB = AN gives AD = 2 * AN, which means N is the midpoint of AD.

Therefore, if the angle bisector of one vertex of a parallelogram bisects the opposite side, it means the side adjacent to that vertex is half the length of the opposite side. Consequently, the angle bisector of the adjacent vertex will bisect the other opposite side for the same reason.

Given:

ABCDE is a regular pentagon.

ABFG is a square.

The shapes are on opposite sides of AB.

To Find:

The measure of $\angle BCF$.

Solution:

In a regular pentagon, the measure of each interior angle is $\frac{(5-2) \times 180^\circ}{5} = \frac{3 \times 180^\circ}{5} = 108^\circ$.

So, $\angle ABC = 108^\circ$.

All sides of a regular pentagon are equal, so BC = AB.

In a square, the measure of each interior angle is $90^\circ$.

So, $\angle ABF = 90^\circ$.

All sides of a square are equal, so BF = AB.

Since BC = AB and BF = AB, we have BC = BF.

Consider $\triangle$BCF. Since BC = BF, $\triangle$BCF is an isosceles triangle with base CF.

The angles opposite the equal sides are equal: $\angle BCF = \angle BFC$.

Now, find the measure of the angle at vertex B in $\triangle$BCF, which is $\angle CBF$.

The angle $\angle ABC$ (interior angle of pentagon) is $108^\circ$.

The angle $\angle ABF$ (interior angle of square) is $90^\circ$.

Since the pentagon and square are on opposite sides of the common side AB, the angle between the segments BC and BF at B is the sum of the angles formed by these segments with AB.

Angle between ray BC and ray BA is $108^\circ$.

Angle between ray BF and ray BA is $90^\circ$.

Since C and F are on opposite sides of the line AB, the angle $\angle CBF$ is formed by rays BC and BF, with ray BA in between them.

Let's find the angle between ray BC and ray BF. Place B at the origin. Let ray BA lie along the positive x-axis. Ray BC makes an angle of $108^\circ$ (say, counterclockwise) from BA. Ray BF makes an angle of $90^\circ$ (clockwise) from BA. The total angle between ray BC and ray BF is $108^\circ + 90^\circ = 198^\circ$. This is the reflex angle. The interior angle $\angle CBF$ of $\triangle$BCF is the smaller angle formed by the rays, which is $360^\circ - 198^\circ = 162^\circ$.

So, $\angle CBF = 162^\circ$.

In $\triangle$BCF, the sum of angles is $180^\circ$.

$\angle CBF + \angle BCF + \angle BFC = 180^\circ$

Substitute $\angle CBF = 162^\circ$ and $\angle BCF = \angle BFC$:

$162^\circ + \angle BCF + \angle BCF = 180^\circ$

$162^\circ + 2 \angle BCF = 180^\circ$

$2 \angle BCF = 180^\circ - 162^\circ$

$2 \angle BCF = 18^\circ$

$\angle BCF = \frac{18^\circ}{2}$

$\angle BCF = 9^\circ$

The measure of $\angle BCF$ is $9^\circ$.

Given:

Convex polygons: quadrilateral, pentagon, hexagon.

To Find:

Maximum number of acute angles and generalization.

Solution:

In a convex polygon, the sum of the exterior angles is always $360^\circ$.

If an interior angle is acute ($< 90^\circ$), its corresponding exterior angle is greater than $90^\circ$ ($180^\circ - \text{acute angle} > 180^\circ - 90^\circ = 90^\circ$).

Let the number of acute interior angles be $k$. The sum of the corresponding $k$ exterior angles is $> k \times 90^\circ$.

Since the sum of all exterior angles is $360^\circ$, we have:

$360^\circ \ge \text{Sum of exterior angles for acute interior angles} > k \times 90^\circ$

$360^\circ > k \times 90^\circ$

Dividing by $90^\circ$, we get $4 > k$.

Thus, the number of acute interior angles ($k$) in a convex polygon must be less than 4.

The maximum possible integer value for $k$ is 3.

Applying this to the given polygons:

1. Convex Quadrilateral ($n=4$): Maximum number of acute angles is 3.

2. Convex Pentagon ($n=5$): Maximum number of acute angles is 3.

3. Convex Hexagon ($n=6$): Maximum number of acute angles is 3.

Pattern Observation and Generalization:

For any convex polygon with $n \ge 4$ sides, the maximum number of acute angles is 3.

(Note: For a convex triangle, $n=3$, the maximum number of acute angles is also 3).

Given:

From the figure and the question, we have:

Line FD is parallel to line BC (FD || BC).

Line BC is parallel to line AE (BC || AE).

Line AC is parallel to line ED (AC || ED).

$\angle ABC = 40^\circ$

$\angle BCD = x$

$\angle CDF = 100^\circ$

To Find:

The value of x.

Solution:

We are given that FD || BC.

Consider the line segment CD as a transversal intersecting the parallel lines FD and BC.

The angles $\angle BCD$ and $\angle CDF$ are interior angles on the same side of the transversal CD.

According to the property of parallel lines intersected by a transversal, the sum of consecutive interior angles is $180^\circ$ (if the lines are parallel).

Therefore, $\angle BCD + \angle CDF = 180^\circ$.

Substitute the given values of these angles:

$x + 100^\circ = 180^\circ$

To find x, subtract $100^\circ$ from both sides of the equation:

$x = 180^\circ - 100^\circ$

$x = 80^\circ$

The value of x is $80^\circ$.

Note: The other given parallel conditions (BC || AE and AC || ED) and the angle $\angle ABC = 40^\circ$ describe other relationships in the figure but are not required to find the value of x based on the consecutive interior angles formed by the transversal CD intersecting the parallel lines FD and BC.

Given:

From the figure, we have a quadrilateral ABCD.

AB || DC.

AD = BC.

$\angle A = 3x + 10^\circ$

$\angle B = 2x + 20^\circ$

To Find:

The value of x.

Solution:

The quadrilateral ABCD is a trapezium since one pair of opposite sides (AB and DC) is parallel.

Since the non-parallel sides are equal in length (AD = BC), the trapezium is an isosceles trapezium.

In an isosceles trapezium, the angles on each parallel base are equal.

Therefore, the angles on the base AB are equal: $\angle DAB = \angle CBA$, which means $\angle A = \angle B$.

Substitute the given expressions for $\angle A$ and $\angle B$:

$3x + 10^\circ = 2x + 20^\circ$

Subtract $2x$ from both sides:

$3x - 2x + 10^\circ = 20^\circ$

$x + 10^\circ = 20^\circ$

Subtract $10^\circ$ from both sides:

$x = 20^\circ - 10^\circ$

$x = 10^\circ$

The value of x is $10^\circ$.

Let's find the angles of the trapezium:

$\angle A = 3x + 10^\circ = 3(10^\circ) + 10^\circ = 30^\circ + 10^\circ = 40^\circ$

$\angle B = 2x + 20^\circ = 2(10^\circ) + 20^\circ = 20^\circ + 20^\circ = 40^\circ$

So, $\angle A = \angle B = 40^\circ$.

In a trapezium with parallel sides AB and DC, consecutive interior angles are supplementary.

$\angle A + \angle D = 180^\circ \implies 40^\circ + \angle D = 180^\circ \implies \angle D = 140^\circ$.

$\angle B + \angle C = 180^\circ \implies 40^\circ + \angle C = 180^\circ \implies \angle C = 140^\circ$.

In an isosceles trapezium, angles on the other parallel base are also equal, i.e., $\angle D = \angle C$. Our result $140^\circ = 140^\circ$ confirms this.

Given:

A trapezium ABCD where AB || DC.

AD = 3 cm

AB = 4 cm

CD = 8 cm

$\angle A = 105^\circ$

To Construct:

Trapezium ABCD.

Steps of Construction:

In a trapezium ABCD with AB || DC, the sum of consecutive interior angles on the same leg is $180^\circ$.

Since AD is a leg connecting the parallel sides AB and DC, $\angle A + \angle D = 180^\circ$.

Given $\angle A = 105^\circ$, we can find $\angle D$:

$\angle D = 180^\circ - \angle A = 180^\circ - 105^\circ = 75^\circ$.

Now we can construct the trapezium:

1. Draw a line segment DC of length 8 cm.

2. At point D, construct an angle of $75^\circ$. Let the ray be DX.

3. On the ray DX, mark point A such that DA = 3 cm.

4. At point A, construct an angle of $105^\circ$ such that the angle is on the same side of AD as DC is. Let this ray be AY. Since $\angle ADC + \angle DAY = 75^\circ + 105^\circ = 180^\circ$, the ray AY is parallel to DC (and thus parallel to AB).

5. On the ray AY, mark point B such that AB = 4 cm.

6. Join points B and C.

7. ABCD is the required trapezium.

Justification:

By construction, DC = 8 cm, AD = 3 cm, and AB = 4 cm.

We constructed $\angle ADC = 75^\circ$ and $\angle DAB = 105^\circ$.

Since $\angle ADC + \angle DAB = 75^\circ + 105^\circ = 180^\circ$, and these are consecutive interior angles with AD as transversal, the line segment AB is parallel to the line segment DC. Thus, AB || DC.

Therefore, ABCD is a trapezium satisfying all the given conditions.

Given:

A parallelogram ABCD with:

AB = 4 cm

BC = 5 cm

$\angle B = 60^\circ$

To Construct:

Parallelogram ABCD.

Steps of Construction:

1. Draw a line segment BC of length 5 cm.

2. At point B, construct an angle of $60^\circ$ using a protractor or compass. Let this ray be BX.

3. On the ray BX, mark a point A such that BA = 4 cm.

4. Now we need to locate point D. Since ABCD is a parallelogram, AD must be parallel to BC and have length 5 cm. Also, CD must be parallel to AB and have length 4 cm.

5. From point A, draw an arc with radius equal to the length of BC (5 cm).

6. From point C, draw an arc with radius equal to the length of AB (4 cm).

7. The intersection point of the arcs drawn in steps 5 and 6 is the required point D.

8. Join AD and CD.

9. ABCD is the required parallelogram.

Justification:

By construction, BC = 5 cm, AB = 4 cm, and $\angle B = 60^\circ$.

Also, by construction, AD = 5 cm and CD = 4 cm.

Since opposite sides are equal in length (AB=CD=4 cm and BC=AD=5 cm), the quadrilateral ABCD is a parallelogram.

Given:

A rhombus with side length 5 cm and one angle measuring $60^\circ$.

To Construct:

The rhombus.