Chapter 8 Quadrilaterals (Additional Questions)

A four-sided polygon is called a Quadrilateral.

The correct option is (B) Quadrilateral.

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

For a quadrilateral, the number of sides is $n=4$.

Therefore, the sum of the interior angles of a quadrilateral is:

$(4-2) \times 180^\circ = 2 \times 180^\circ = 360^\circ$

The correct option is (B) $360^\circ$.

Let the angles of the quadrilateral be $x$, $2x$, $3x$, and $4x$.

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

Therefore, we have the equation:

$x + 2x + 3x + 4x = 360^\circ$

Combining the terms on the left side:

$10x = 360^\circ$

Solving for $x$:

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

$x = 36^\circ$

Now, we find the measure of each angle:

First angle = $x = 36^\circ$

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

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

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

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

The largest angle among these is $144^\circ$.

The correct option is (D) $144^\circ$.

A parallelogram is a quadrilateral with two pairs of parallel sides.

A fundamental property of parallelograms is that their opposite sides are both equal in length and parallel to each other.

The correct option is (A) Equal and parallel.

One of the key properties of a parallelogram is that its diagonals intersect at their midpoints. This means that each diagonal is divided into two equal segments by the point of intersection.

Therefore, the diagonals of a parallelogram bisect each other.

Options (A), (C), and (D) are properties that are true for specific types of parallelograms (rectangles, rhombuses, squares) but not for all parallelograms.

The correct option is (B) Bisect each other.

A quadrilateral with at least one pair of parallel opposite sides is called a trapezium (or trapezoid).

If the definition specifies exactly one pair of parallel opposite sides, it specifically refers to a trapezium that is not a parallelogram.

Based on common geometric definitions, the figure described is a Trapezium.

The correct option is (C) Trapezium.

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

As a parallelogram, its diagonals bisect each other.

A specific property of a rectangle (which is not true for all parallelograms) is that its diagonals are equal in length.

The diagonals of a rectangle are perpendicular only if the rectangle is also a rhombus (i.e., a square).

Therefore, in a rectangle, the diagonals are equal and bisect each other, but they are not necessarily perpendicular.

The correct option is (B) Equal and bisect each other (not necessarily at $90^\circ$).

A rhombus is a quadrilateral with four equal sides. It is also a type of parallelogram because its opposite sides are parallel.

In a general parallelogram, opposite sides are equal. For it to be a rhombus, the adjacent sides must also be equal. If one pair of adjacent sides is equal, then all four sides are equal because opposite sides are already equal in a parallelogram.

Therefore, a rhombus is a parallelogram where adjacent sides are equal.

Option (A) describes a rectangle (or square), not necessarily a rhombus.

Option (B) describes a rectangle (or square), not necessarily a rhombus.

Option (D) also describes a rectangle (or square).

The correct option is (C) Adjacent sides are equal.

A rhombus is a parallelogram with all four sides equal.

As a parallelogram, its diagonals bisect each other.

In addition to bisecting each other, the diagonals of a rhombus have two more special properties:

1. They are perpendicular to each other (bisect each other at $90^\circ$).

2. They bisect the angles of the rhombus.

Option (A) is incorrect because the diagonals of a rhombus are only equal if it is a square.

Option (B) states that they bisect each other at $90^\circ$, which is true.

Option (C) states that they bisect the angles, which is also true.

Option (D) combines both (B) and (C), and since both statements are true for a rhombus, option (D) is the most complete and correct answer.

The correct option is (D) Both (B) and (C).

A square is defined as a quadrilateral with all four sides equal in length and all four interior angles equal to $90^\circ$.

A Rectangle is a quadrilateral with four right angles.

A Rhombus is a quadrilateral with four equal sides.

Since a square has four right angles, it satisfies the definition of a rectangle.

Since a square has four equal sides, it satisfies the definition of a rhombus.

Therefore, a square is a quadrilateral that is simultaneously a Rectangle and a Rhombus.

The correct option is (B) Rectangle, Rhombus.

Given that M, N, and P are the mid-points of sides AB, BC, and CA respectively of $\triangle ABC$.

We will use the Midpoint Theorem.

The Midpoint Theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.

Considering $\triangle ABC$:

Since M is the midpoint of AB and P is the midpoint of AC, by the Midpoint Theorem, we have:

$MP \parallel BC$ and $MP = \frac{1}{2} BC$

Since N is the midpoint of BC, BN is part of BC. Therefore, $MP \parallel BN$.

Also, consider the segment NP.

Since N is the midpoint of BC and P is the midpoint of AC, by the Midpoint Theorem, we have:

$NP \parallel AB$ and $NP = \frac{1}{2} AB$

Since M is the midpoint of AB, AM is part of AB. Therefore, $NP \parallel AM$.

Now consider the quadrilateral AMNP. The vertices are A, M, N, P.

The sides of the quadrilateral AMNP are AM, MN, NP, and PA.

We have $NP \parallel AM$ (since $NP \parallel AB$ and M lies on AB).

We have $MN \parallel AP$ (since M is midpoint of AB, N is midpoint of BC, by Midpoint Theorem $MN \parallel AC$, and P lies on AC).

A quadrilateral with both pairs of opposite sides parallel is a Parallelogram.

Additionally, we can also show that opposite sides are equal:

$AM = \frac{1}{2} AB$ (since M is the midpoint of AB)

$NP = \frac{1}{2} AB$ (by Midpoint Theorem for $\triangle ABC$ and points N, P)

Thus, $AM = NP$.

$AP = \frac{1}{2} AC$ (since P is the midpoint of AC)

$MN = \frac{1}{2} AC$ (by Midpoint Theorem for $\triangle ABC$ and points M, N)

Thus, $AP = MN$.

Since both pairs of opposite sides are equal and parallel, AMNP is a Parallelogram.

The correct option is (B) Parallelogram.

The Mid-point Theorem states that the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of its length.

Based on this theorem, the blanks should be filled with "parallel" and "half".

The correct option is (B) Parallel, half.

Let's analyze each statement:

(A) Every parallelogram is a quadrilateral. A parallelogram is a polygon with four sides, which is the definition of a quadrilateral. This statement is TRUE.

(B) Every rectangle is a parallelogram. A rectangle is a quadrilateral with four right angles. This implies that opposite sides are parallel (lines perpendicular to the same line are parallel), so a rectangle is a parallelogram. This statement is TRUE.

(C) Every rhombus is a square. A rhombus is a quadrilateral with four equal sides. A square is a rhombus with four right angles. Not all rhombuses have four right angles. For example, a rhombus can have angles of $60^\circ$ and $120^\circ$. Therefore, not every rhombus is a square. This statement is FALSE.

(D) Every square is a rhombus and a rectangle. A square has four equal sides (property of a rhombus) and four right angles (property of a rectangle). Therefore, a square is both a rhombus and a rectangle. This statement is TRUE.

The statement that is FALSE is (C) Every rhombus is a square.

The correct option is (C) Every rhombus is a square.

In a parallelogram, consecutive angles are supplementary, meaning their sum is $180^\circ$.

Given that ABCD is a parallelogram, $\angle A$ and $\angle B$ are consecutive angles.

Therefore, we have:

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

We are given $\angle A = 75^\circ$. Substitute this value into the equation:

$75^\circ + \angle B = 180^\circ$

To find $\angle B$, subtract $75^\circ$ from both sides:

$\angle B = 180^\circ - 75^\circ$

$\angle B = 105^\circ$

The measure of $\angle B$ is $105^\circ$.

The correct option is (B) $105^\circ$.

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): The diagonals of a rectangle are equal.

This statement is a fundamental property of rectangles. The diagonals of a rectangle always have the same length. So, Assertion (A) is True.

Reason (R): A rectangle is a parallelogram with one angle equal to $90^\circ$.

This is a valid definition of a rectangle. If a parallelogram has one angle of $90^\circ$, then due to the properties of parallelograms (opposite angles are equal, consecutive angles are supplementary), all four angles must be $90^\circ$. Thus, Reason (R) is True.

Now, let's consider if Reason (R) is the correct explanation for Assertion (A).

The fact that a rectangle is a parallelogram with a right angle is precisely the property used to prove that its diagonals are equal. Consider a rectangle ABCD. Since it's a parallelogram, AB is parallel to DC and AD is parallel to BC. Also, suppose $\angle A = 90^\circ$. Then all angles are $90^\circ$.

Consider $\triangle ADC$ and $\triangle BCD$.

AD = BC (Opposite sides of a parallelogram are equal)

DC = DC (Common side)

$\angle ADC = \angle BCD = 90^\circ$ (Angles of a rectangle)

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

Therefore, AC = BD (Corresponding parts of congruent triangles are equal).

The proof relies on the properties stated or implied by Reason (R). Thus, Reason (R) provides the essential characteristic of a rectangle that leads to the equality of its diagonals.

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): The diagonals of a rhombus are perpendicular bisectors of each other.

This statement is a property of a rhombus. The diagonals of a rhombus bisect each other (because a rhombus is a parallelogram) and they intersect at a $90^\circ$ angle, making them perpendicular bisectors. So, Assertion (A) is True.

Reason (R): A rhombus is a parallelogram with all sides equal.

This is a valid definition of a rhombus. A rhombus is indeed a parallelogram where all four sides are of equal length. So, Reason (R) is True.

Now, let's consider if Reason (R) is the correct explanation for Assertion (A).

The property that a rhombus has all sides equal (as stated in R) is the key characteristic that distinguishes it from a general parallelogram and leads to the property that its diagonals are perpendicular. In a parallelogram, diagonals only bisect each other. The additional property of perpendicularity arises because all sides are equal. Consider a rhombus ABCD with diagonals intersecting at O. Since it's a parallelogram, the diagonals bisect each other, so AO = OC and BO = OD. Now, consider $\triangle AOB$ and $\triangle COB$. AB = CB (sides of a rhombus are equal), AO = CO (diagonals bisect each other), and BO is common. By SSS congruence, $\triangle AOB \cong \triangle COB$. Therefore, $\angle AOB = \angle COB$. Since $\angle AOB$ and $\angle COB$ form a linear pair, their sum is $180^\circ$. Thus, $\angle AOB = \angle COB = \frac{180^\circ}{2} = 90^\circ$. This proves that the diagonals are perpendicular. The property mentioned in R (all sides equal) is used to prove the perpendicularity of the bisecting diagonals.

Therefore, Reason (R) is the correct explanation for Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Let's match each quadrilateral with the correct property of its diagonals:

(i) Parallelogram: The diagonals of a parallelogram bisect each other. This matches property (b).

(ii) Rectangle: A rectangle is a parallelogram with equal diagonals. Thus, the diagonals are equal and bisect each other. This matches property (c).

(iii) Rhombus: A rhombus is a parallelogram whose diagonals bisect each other at right angles. This matches property (d).

(iv) Square: A square is both a rectangle and a rhombus. Its diagonals have the properties of both: they are equal (like a rectangle) and they bisect each other at right angles (like a rhombus). This matches property (a).

The correct matching is:

(i) - (b)

(ii) - (c)

(iii) - (d)

(iv) - (a)

Comparing this matching with the given options, we find that option (A) corresponds to this matching.

The correct option is (A) (i)-(b), (ii)-(c), (iii)-(d), (iv)-(a).

Let's analyze the properties of the diagonals of a rectangle ABCD, which intersect at point O.

A rectangle is a type of parallelogram.

Property 1: The diagonals of a parallelogram bisect each other. This means that the intersection point O is the midpoint of both diagonals AC and BD.

From this property, we have $AO = OC$ and $BO = OD$.

Property 2: The diagonals of a rectangle are equal in length.

So, $AC = BD$.

Now let's evaluate the given statements based on these properties:

(A) $AO = OC$ but $BO \neq OD$. This is false because, as a parallelogram, both diagonals bisect each other, so $BO$ must be equal to $OD$.

(B) $AO = BO = CO = DO$. Since O is the midpoint of AC and BD, $AO = \frac{1}{2} AC$ and $BO = \frac{1}{2} BD$. Because the diagonals of a rectangle are equal ($AC = BD$), their halves are also equal. Thus, $\frac{1}{2} AC = \frac{1}{2} BD$, which means $AO = BO$. Combining this with the bisection property ($AO=OC$ and $BO=OD$), we get $AO = BO = CO = DO$. This statement is TRUE.

(C) The angle $\angle AOB$ is always $90^\circ$. The diagonals of a parallelogram are perpendicular only if it is a rhombus. A rectangle is a rhombus only if it is a square. So, $\angle AOB = 90^\circ$ only if the rectangle is a square. For a general rectangle, the angle is not $90^\circ$. This statement is false.

(D) The diagonal AC bisects $\angle BAD$. A diagonal bisects the angle of a vertex only in a rhombus or a square. In a general rectangle that is not a square, the diagonal does not bisect the vertex angle (which is $90^\circ$). For example, if the sides are of different lengths, the angles formed by the diagonal will be different. This statement is false.

Based on the analysis, the only true statement is (B).

The correct option is (B) $AO = BO = CO = DO$.

Given that in $\triangle PQR$, S is the mid-point of side PQ and T is the mid-point of side PR.

We are given the length of the segment ST as $ST = 5$ cm.

We can apply the Midpoint Theorem here.

The Midpoint Theorem states that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.

In $\triangle PQR$, the segment ST joins the midpoints S and T of sides PQ and PR, respectively.

According to the Midpoint Theorem, ST is parallel to the third side QR, and the length of ST is half the length of QR.

So, we have the relationship:

$ST = \frac{1}{2} QR$

We are given $ST = 5$ cm. Substitute this value into the equation:

$5 = \frac{1}{2} QR$

To find QR, multiply both sides of the equation by 2:

$QR = 2 \times 5$

$QR = 10$ cm

The length of the side QR is 10 cm.

The correct option is (B) 10 cm.

Let's analyze the properties of the diagonals given for the quadrilateral:

1. The diagonals bisect each other: This property holds for all Parallelograms.

2. The diagonals are equal: A parallelogram with equal diagonals is a Rectangle.

3. The diagonals bisect each other at right angles: A parallelogram whose diagonals bisect each other at right angles is a Rhombus.

The question states that the diagonals of the quadrilateral have all three properties:

- They bisect each other (Property of a parallelogram).

- They are equal (Additional property of a rectangle).

- They bisect each other at right angles (Additional property of a rhombus).

Therefore, the quadrilateral must be a parallelogram that is both a rectangle and a rhombus.

A quadrilateral that is both a rectangle (all angles are $90^\circ$) and a rhombus (all sides are equal) is a Square.

A square has all the mentioned properties for its diagonals:

- Diagonals bisect each other (since it's a parallelogram).

- Diagonals are equal (since it's a rectangle).

- Diagonals are perpendicular bisectors of each other (since it's a rhombus).

Let's check the options:

(A) Rectangle: Diagonals are equal and bisect each other, but not necessarily at right angles.

(B) Rhombus: Diagonals bisect each other at right angles, but not necessarily equal.

(C) Square: Diagonals are equal and bisect each other at right angles. This matches all given properties.

(D) Parallelogram: Diagonals only necessarily bisect each other.

Thus, the quadrilateral is a square.

The correct option is (C) Square.

Let's examine each statement to determine if the described quadrilateral must be a parallelogram.

(A) A quadrilateral with opposite sides parallel.

This is the standard definition of a parallelogram. A quadrilateral is a parallelogram if and only if both pairs of opposite sides are parallel.

This statement must be a parallelogram.

(B) A quadrilateral with opposite angles equal.

Let the quadrilateral be ABCD. If $\angle A = \angle C$ and $\angle B = \angle D$.

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

$\angle A + \angle B + \angle C + \angle D = 360^\circ$

Substitute $\angle C = \angle A$ and $\angle D = \angle B$:

$\angle A + \angle B + \angle A + \angle B = 360^\circ$

$2(\angle A + \angle B) = 360^\circ$

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

Since consecutive angles ($\angle A$ and $\angle B$) are supplementary, the sides AB and CD must be parallel, and AD and BC must be parallel.

Thus, a quadrilateral with opposite angles equal must be a parallelogram.

(C) A quadrilateral with adjacent sides equal.

Consider a kite. A kite is a quadrilateral with two pairs of equal-length sides that are adjacent to each other. For example, in quadrilateral ABCD, AB = BC and AD = CD. A kite is not necessarily a parallelogram (unless all four sides are equal, which makes it a rhombus, a specific type of parallelogram).

For example, a kite with vertices (0,1), (1,0), (0,-2), (-1,0) has adjacent sides equal but is not a parallelogram.

This statement does not necessarily have to be a parallelogram.

(D) A quadrilateral with diagonals bisecting each other.

Let the diagonals AC and BD intersect at O, such that AO = OC and BO = OD.

Consider $\triangle AOB$ and $\triangle COD$.

$AO = CO$ (Given)

$BO = DO$ (Given)

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

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

Therefore, AB = CD (Corresponding parts of congruent triangles) and $\angle OAB = \angle OCD$ (Alternate interior angles if AB || CD).

Since $\angle OAB = \angle OCD$, this implies that line AB is parallel to line CD (as they are alternate interior angles formed by transversal AC). Similarly, considering $\triangle AOD$ and $\triangle COB$, we can prove AD || BC and AD = BC.

Thus, a quadrilateral with diagonals bisecting each other must be a parallelogram.

Based on the analysis, statements (A), (B), and (D) describe properties that guarantee a quadrilateral is a parallelogram.

The statements that must be a parallelogram are (A), (B), and (D).

In a parallelogram, opposite angles are equal.

Given that ABCD is a parallelogram, $\angle A$ and $\angle C$ are opposite angles.

Therefore, we have:

$\angle A = \angle C$

Substitute the given expressions for $\angle A$ and $\angle C$:

$(2x+10)^\circ = (3x-20)^\circ$

Now, we solve for $x$. We can ignore the degree symbols during calculation.

$2x + 10 = 3x - 20$

Subtract $2x$ from both sides:

$10 = 3x - 20 - 2x$

$10 = x - 20$

Add 20 to both sides:

$10 + 20 = x$

$30 = x$

So, the value of $x$ is 30.

We can check the angle measures:

$\angle A = (2 \times 30 + 10)^\circ = (60 + 10)^\circ = 70^\circ$

$\angle C = (3 \times 30 - 20)^\circ = (90 - 20)^\circ = 70^\circ$

Since $\angle A = \angle C = 70^\circ$, our value of $x$ is correct.

The value of $x$ is 30.

The correct option is (C) $30^\circ$.

The Mid-point Theorem states: "The line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side."

The converse of the Mid-point Theorem states: "A line drawn through the midpoint of one side of a triangle, parallel to another side, bisects the third side."

The statement given in the question is: "The line segment joining the mid-points of one side of a triangle, parallel to another side, bisects the third side." This statement matches the wording of the converse of the Mid-point Theorem.

Therefore, the given statement is the converse of the Mid-point Theorem.

The correct option is (B) Mid-point Theorem.

Let's analyze the given conditions for the diagonals of the quadrilateral:

1. The diagonals are equal.

2. The diagonals do not bisect each other at right angles.

Consider the properties of the diagonals for each option:

(A) Square: The diagonals are equal, bisect each other, and bisect each other at right angles ($\mathbf{90^\circ}$). This quadrilateral satisfies the first condition (diagonals are equal) but fails the second condition (they *do* bisect each other at right angles).

(B) Rectangle (not a square): A rectangle is a parallelogram with equal diagonals. The diagonals of a parallelogram bisect each other. Thus, the diagonals of a rectangle are equal and bisect each other. In a rectangle that is not a square, the diagonals intersect at an angle that is not $90^\circ$. This quadrilateral satisfies both conditions: diagonals are equal and they bisect each other but not at right angles. Note that "do not bisect each other at right angles" implies that bisection occurs, but not perpendicularly.

(C) Rhombus (not a square): The diagonals bisect each other at right angles. However, the diagonals of a rhombus are equal only if it is a square. For a rhombus that is not a square, the diagonals are not equal. This quadrilateral fails the first condition (diagonals are equal).

(D) Isosceles Trapezium: The diagonals of an isosceles trapezium are equal. However, the diagonals of an isosceles trapezium do not bisect each other (unless it is also a rectangle). If the diagonals do not bisect each other, the statement "bisect each other at right angles" is false. The statement "do not bisect each other at right angles" is true in this case because the bisection part is false. While this option satisfies the condition that diagonals are equal, the phrasing "do not bisect each other at right angles" typically refers to figures where bisection *does* occur but is not perpendicular.

Comparing the options with the given conditions, a Rectangle (not a square) is the quadrilateral whose diagonals are equal and bisect each other at an angle other than $90^\circ$. This perfectly matches the conditions provided in the question.

The correct option is (B) Rectangle (not a square).

Let's analyze each statement in the context of the properties of a parallelogram.

(A) Opposite sides are equal. This is a defining property of a parallelogram. If a quadrilateral is a parallelogram, its opposite sides are equal in length. Conversely, if opposite sides are equal, the quadrilateral is a parallelogram.

This is a property of a parallelogram.

(B) Opposite angles are equal. This is also a property of a parallelogram. If a quadrilateral is a parallelogram, its opposite angles are equal. Conversely, if opposite angles are equal, the quadrilateral is a parallelogram.

This is a property of a parallelogram.

(C) Diagonals are perpendicular. The diagonals of a parallelogram are perpendicular if and only if the parallelogram is a rhombus. A rhombus is a specific type of parallelogram, but not all parallelograms are rhombuses (e.g., a rectangle that is not a square is a parallelogram, but its diagonals are not perpendicular). Therefore, this is not a property that holds for *all* parallelograms.

This is NOT a property of a parallelogram in general.

(D) Consecutive angles are supplementary. This is a property of a parallelogram. In a parallelogram, adjacent angles sum up to $180^\circ$. For example, in parallelogram ABCD, $\angle A + \angle B = 180^\circ$, $\angle B + \angle C = 180^\circ$, etc. This property arises because opposite sides are parallel.

This is a property of a parallelogram.

The statement that is NOT a property of a parallelogram is that its diagonals are perpendicular.

The correct option is (C) Diagonals are perpendicular.

We are given a quadrilateral that is already a parallelogram.

We are given the additional condition that its diagonals are equal.

Let's consider the properties of the diagonals for different types of parallelograms:

- In a general parallelogram, diagonals bisect each other.

- In a Rectangle, the diagonals are equal and bisect each other.

- In a Rhombus, the diagonals bisect each other at right angles and bisect the angles (but are not necessarily equal).

- In a Square, the diagonals are equal, bisect each other at right angles, and bisect the angles.

A parallelogram with equal diagonals is specifically a Rectangle.

A square is a type of rectangle (one whose adjacent sides are equal), and its diagonals are also equal. However, the property of having equal diagonals is what elevates a parallelogram to the status of a rectangle, not necessarily a square unless other conditions (like perpendicular diagonals) are met.

A rhombus does not generally have equal diagonals; they are equal only if the rhombus is also a square.

A trapezium is not necessarily a parallelogram.

Therefore, if the diagonals of a parallelogram are equal, it must be a rectangle.

The correct option is (C) Rectangle.

Given that in $\triangle LMN$, P is the mid-point of side LM and Q is the mid-point of side LN.

We are given the length of the segment PQ as $PQ = 8$ cm.

We can apply the Midpoint Theorem here.

The Midpoint Theorem states that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.

In $\triangle LMN$, the segment PQ joins the midpoints P and Q of sides LM and LN, respectively.

According to the Midpoint Theorem, PQ is parallel to the third side MN, and the length of PQ is half the length of MN.

So, we have the relationship:

$PQ = \frac{1}{2} MN$

We are given $PQ = 8$ cm. Substitute this value into the equation:

$8 = \frac{1}{2} MN$

To find MN, multiply both sides of the equation by 2:

$MN = 2 \times 8$

$MN = 16$ cm

The length of the side MN is 16 cm.

The correct option is (C) 16 cm.

A rhombus is a special type of parallelogram (one where all sides are equal).

One of the fundamental properties of a parallelogram is that its diagonals bisect each other.

To bisect means to divide into two equal parts.

Since a rhombus is a parallelogram, its diagonals also bisect each other.

Therefore, the point of intersection of the diagonals of a rhombus divides each diagonal into two equal parts.

While the diagonals of a rhombus also intersect at right angles (are perpendicular) and bisect the angles of the rhombus, the question specifically asks how the intersection point divides the diagonals, which relates to bisection.

The correct option is (C) Equal.

Let's analyze the properties of the diagonals given for the quadrilateral:

1. The diagonals bisect each other.

2. The diagonals intersect at right angles ($90^\circ$).

Consider the first property: The diagonals bisect each other. A quadrilateral whose diagonals bisect each other is always a Parallelogram.

Now, consider the second property in addition to the first: The diagonals of this parallelogram also intersect at right angles.

A parallelogram whose diagonals are perpendicular is a Rhombus.

Let's check the options:

(A) Parallelogram: Diagonals bisect each other, but not necessarily at right angles.

(B) Rectangle: Diagonals are equal and bisect each other, but not necessarily at right angles.

(C) Rhombus: Diagonals bisect each other (because it's a parallelogram) and intersect at right angles. This matches the given properties.

(D) Trapezium: Diagonals do not necessarily bisect each other, nor are they necessarily perpendicular.

Therefore, if the diagonals of a quadrilateral bisect each other at right angles, it is a rhombus.

The correct option is (C) Rhombus.

Given that the four towns A, B, C, and D form a parallelogram ABCD when connected by straight roads.

We are given the distances between some towns:

AB = 10 km

BC = 6 km

We need to find the distance between town C and town D, which is the length of the side CD.

One of the fundamental properties of a parallelogram is that its opposite sides are equal in length.

In parallelogram ABCD, the pairs of opposite sides are (AB and CD) and (BC and AD).

Therefore, the length of side AB is equal to the length of side CD.

AB = CD

Similarly, the length of side BC is equal to the length of side AD (AD = BC = 6 km), although this is not required to answer the question.

The distance between town C and town D is 10 km.

The correct option is (A) 10 km.

Let the angles of the quadrilateral be $\angle 1 = 90^\circ$, $\angle 2 = 70^\circ$, $\angle 3 = 110^\circ$, and $\angle 4 = x^\circ$.

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

Therefore, we can write the equation:

$\angle 1 + \angle 2 + \angle 3 + \angle 4 = 360^\circ$

Substitute the given values into the equation:

$90^\circ + 70^\circ + 110^\circ + x^\circ = 360^\circ$

Sum the known angles:

$(90 + 70 + 110)^\circ + x^\circ = 360^\circ$

$(160 + 110)^\circ + x^\circ = 360^\circ$

$270^\circ + x^\circ = 360^\circ$

Subtract $270^\circ$ from both sides to solve for $x$:

$x^\circ = 360^\circ - 270^\circ$

$x^\circ = 90^\circ$

The value of $x$ is 90.

The correct option is (A) $90^\circ$.

Let's analyze the given property of the diagonals of the quadrilateral:

1. The diagonals bisect each other.

2. The diagonals are perpendicular to each other.

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

If the diagonals of a parallelogram are perpendicular, the parallelogram is a Rhombus.

Therefore, a quadrilateral whose diagonals are perpendicular bisectors of each other is a Rhombus.

Now let's consider which of the given options must be true if a quadrilateral has diagonals that are perpendicular bisectors of each other:

(A) Parallelogram: Since the diagonals bisect each other, the quadrilateral must be a parallelogram. So, this is true.

(B) Rhombus: A rhombus is defined as a parallelogram with four equal sides, or equivalently, a quadrilateral whose diagonals are perpendicular bisectors of each other. This is true by definition based on the given property.

(C) Square: A square is a rhombus with four right angles (or equal diagonals). While the diagonals of a square are perpendicular bisectors, not every quadrilateral with perpendicular bisecting diagonals is a square (it is a square only if the diagonals are also equal). So, this is not necessarily true.

(D) Kite: A kite is a quadrilateral with perpendicular diagonals where one diagonal bisects the other. In the case where the diagonals are perpendicular *bisectors* (meaning *both* diagonals are bisected), the quadrilateral is a rhombus. A rhombus is a special type of kite where adjacent sides are equal. So, a rhombus is a kite. Thus, this is true.

Based on the property that the diagonals are perpendicular bisectors, the quadrilateral must be a Rhombus. Since every Rhombus is also a Parallelogram and a Kite, the quadrilateral belongs to these categories as well.

The correct options are (A) Parallelogram, (B) Rhombus, and (D) Kite.

Let the rhombus be ABCD, and let P, Q, R, and S be the midpoints of sides AB, BC, CD, and DA respectively. We need to determine the type of quadrilateral PQRS.

Consider $\triangle ABC$. Since P and Q are the midpoints of AB and BC, by the Midpoint Theorem:

Consider $\triangle ADC$. Since R and S are the midpoints of CD and DA, by the Midpoint Theorem:

From (i) and (iii), we have PQ $\parallel$ AC and SR $\parallel$ AC, which implies PQ $\parallel$ SR.

From (ii) and (iv), we have PQ = $\frac{1}{2}$ AC and SR = $\frac{1}{2}$ AC, which implies PQ = SR.

Since one pair of opposite sides (PQ and SR) is both equal and parallel, the quadrilateral PQRS is a Parallelogram.

Now let's consider the other pair of sides, PS and QR.

Consider $\triangle ABD$. Since P and S are the midpoints of AB and AD, by the Midpoint Theorem:

PS $\parallel$ BD

PS = $\frac{1}{2}$ BD

Consider $\triangle BCD$. Since Q and R are the midpoints of BC and CD, by the Midpoint Theorem:

QR $\parallel$ BD

QR = $\frac{1}{2}$ BD

So, PS $\parallel$ QR and PS = QR. This further confirms that PQRS is a parallelogram.

Now, consider the diagonals of the rhombus, AC and BD. A key property of a rhombus is that its diagonals are perpendicular to each other.

AC $\perp$ BD

We know that PQ $\parallel$ AC and PS $\parallel$ BD.

If two lines are parallel to two perpendicular lines, then the first two lines are also perpendicular to each other.

Since PQ is parallel to AC and PS is parallel to BD, and AC is perpendicular to BD, it follows that PQ is perpendicular to PS.

Therefore, the angle between adjacent sides PQ and PS in the parallelogram PQRS is $90^\circ$, i.e., $\angle SPQ = 90^\circ$.

A parallelogram with one angle equal to $90^\circ$ is a Rectangle.

To check if it's a square, we need to see if adjacent sides are equal, i.e., if PQ = PS.

PQ = $\frac{1}{2}$ AC and PS = $\frac{1}{2}$ BD.

For PQ = PS, we would need $\frac{1}{2}$ AC = $\frac{1}{2}$ BD, which means AC = BD. The diagonals of a rhombus are equal only if the rhombus is a square. Since the problem specifies a general rhombus (not necessarily a square), the diagonals AC and BD are not necessarily equal. Therefore, PQ is not necessarily equal to PS.

Thus, the figure is a rectangle, but not necessarily a square.

The correct option is (C) Rectangle.

Let's analyze each condition to determine which one guarantees that a quadrilateral ABCD is a parallelogram.

(A) $AB \parallel CD$

This condition means that only one pair of opposite sides is parallel. A quadrilateral with at least one pair of parallel opposite sides is called a trapezium. A trapezium is not necessarily a parallelogram (unless the other pair of opposite sides is also parallel).

This condition alone is not sufficient for a parallelogram.

(B) $AB = CD$

This condition means that only one pair of opposite sides is equal in length. Consider an isosceles trapezium where the non-parallel sides are equal, and the parallel sides are also equal. Such a figure is not necessarily a parallelogram. A quadrilateral with only one pair of opposite sides equal is not guaranteed to be a parallelogram.

This condition alone is not sufficient for a parallelogram.

(C) Opposite angles are supplementary.

Let the angles be $\angle A, \angle B, \angle C, \angle D$. If $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$. The sum of all angles in a quadrilateral is $360^\circ$. $\angle A + \angle B + \angle C + \angle D = (\angle A + \angle C) + (\angle B + \angle D) = 180^\circ + 180^\circ = 360^\circ$. This condition is always true for any quadrilateral, as long as the sum of opposite angles is $180^\circ$. This does not uniquely define a parallelogram. For example, an isosceles trapezium that is not a rectangle has opposite angles supplementary in pairs, but it's not a parallelogram.

This condition alone is not sufficient for a parallelogram.

(D) Diagonals bisect each other.

Let the diagonals AC and BD intersect at point O. If the diagonals bisect each other, then $AO = OC$ and $BO = OD$. This is a defining property of a parallelogram. It can be proven that if the diagonals of a quadrilateral bisect each other, then its opposite sides are parallel, making it a parallelogram.

This condition is sufficient for a parallelogram.

Therefore, a quadrilateral ABCD is a parallelogram if its diagonals bisect each other.

The correct option is (D) Diagonals bisect each other.

The question describes the scenario and explicitly refers to the converse of the Mid-point Theorem.

Let's recall the statement of the Converse of the Mid-point Theorem:

"A line drawn through the midpoint of one side of a triangle, parallel to another side, bisects the third side."

In the given problem:

The triangle is $\triangle XYZ$.

P is the midpoint of side XY (This corresponds to "midpoint of one side").

A line is drawn through P.

This line is parallel to YZ (This corresponds to "parallel to another side").

This line intersects the third side, XZ, at point Q.

According to the converse of the Mid-point Theorem, the line bisects the third side (XZ).

Bisecting the side XZ means dividing it into two equal parts, which implies that Q is the point that divides XZ exactly in half.

Therefore, Q is the mid-point of XZ.

Comparing this conclusion with the given options:

(A) Mid-point - This matches our conclusion.

(B) End-point - Q lies on the side XZ, but is not an end-point (X or Z), unless in a degenerate triangle.

(C) Vertex - Q lies on the side XZ, not a vertex (X, Y, or Z).

(D) Centroid - The centroid is the intersection of medians, not directly related to this construction based on the converse of the midpoint theorem.

Thus, Q is the mid-point of XZ.

The correct option is (A) Mid-point.

We are given a quadrilateral that is already a parallelogram.

We are given the additional condition that its diagonals are perpendicular.

Let's consider the properties of the diagonals for different types of parallelograms:

- In a general parallelogram, diagonals bisect each other.

- In a Rectangle, the diagonals are equal and bisect each other (not necessarily perpendicular).

- In a Rhombus, the diagonals bisect each other at right angles ($\mathbf{90^\circ}$) and bisect the angles.

- In a Square, the diagonals are equal and bisect each other at right angles and bisect the angles.

A parallelogram whose diagonals are perpendicular is specifically a Rhombus.

A square is a type of rhombus (one whose angles are $90^\circ$), and its diagonals are also perpendicular. However, the property of having perpendicular diagonals is what elevates a parallelogram to the status of a rhombus, not necessarily a square unless other conditions (like equal diagonals) are met.

A rectangle does not generally have perpendicular diagonals; they are perpendicular only if the rectangle is also a square.

A trapezium is not necessarily a parallelogram.

Therefore, if the diagonals of a parallelogram are perpendicular, it must be a rhombus.

The correct option is (C) Rhombus.

Let the given rectangle be ABCD. Let P, Q, R, and S be the mid-points of sides AB, BC, CD, and DA respectively.

We want to determine the type of quadrilateral formed by joining these mid-points, which is PQRS.

Consider $\triangle ABC$. Since P and Q are the mid-points of AB and BC, by the Midpoint Theorem:

Consider $\triangle ADC$. Since R and S are the mid-points of CD and DA, by the Midpoint Theorem:

From (i) and (iii), we have PQ $\parallel$ AC and SR $\parallel$ AC, which implies PQ $\parallel$ SR.

From (ii) and (iv), we have PQ = $\frac{1}{2}$ AC and SR = $\frac{1}{2}$ AC, which implies PQ = SR.

Since one pair of opposite sides (PQ and SR) is equal and parallel, quadrilateral PQRS is a Parallelogram.

Now consider $\triangle ABD$. Since P and S are the mid-points of AB and AD, by the Midpoint Theorem:

We know that the diagonals of a rectangle are equal.

From (ii), (vi), and AC = BD, we have:

PQ = $\frac{1}{2}$ AC

PS = $\frac{1}{2}$ BD

Since AC = BD, it follows that $\frac{1}{2}$ AC = $\frac{1}{2}$ BD, so PQ = PS.

We have shown that PQRS is a parallelogram (PQ $\parallel$ SR, PS $\parallel$ QR, PQ = SR, PS = QR).

We have also shown that adjacent sides are equal (PQ = PS).

A parallelogram with equal adjacent sides is a Rhombus.

To check if it is a square, we need to determine if the angles are $90^\circ$. The angle between PQ and PS is the same as the angle between the lines they are parallel to, which are AC and BD respectively. The diagonals of a rectangle are perpendicular only if the rectangle is a square. For a rectangle that is not a square, the diagonals intersect at an angle other than $90^\circ$. Thus, the angles of PQRS are generally not $90^\circ$.

Therefore, the figure formed is a rhombus.

The correct option is (C) Rhombus.

In a parallelogram, opposite angles are equal.

Given that ABCD is a parallelogram, $\angle B$ and $\angle D$ are opposite angles.

Therefore, we have:

$\angle B = \angle D$

We are given $\angle B = 110^\circ$.

So, $\angle D = 110^\circ$.

The measure of $\angle D$ is $110^\circ$.

The correct option is (B) $110^\circ$.

Let's analyze the properties given for the quadrilateral:

1. All sides are equal.

2. All angles are equal.

A quadrilateral with all sides equal is a Rhombus.

A quadrilateral with all angles equal must have each angle equal to $\frac{360^\circ}{4} = 90^\circ$. A quadrilateral with all angles equal to $90^\circ$ is a Rectangle.

The quadrilateral described has both properties: all sides equal (like a rhombus) and all angles equal (like a rectangle, specifically $90^\circ$).

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

Let's check the options:

(A) Rectangle: Has all angles equal, but not necessarily all sides equal.

(B) Rhombus: Has all sides equal, but not necessarily all angles equal.

(C) Square: Has all sides equal and all angles equal. This matches the description.

(D) Parallelogram: Has opposite sides equal and opposite angles equal, but not necessarily all sides or all angles equal.

Thus, a quadrilateral with all sides equal and all angles equal is a square.

The correct option is (C) Square.

Let the rhombus be ABCD and let the swing set be located at the intersection point of the diagonals, O.

The lengths of the diagonals are given as $d_1 = 12$ metres and $d_2 = 16$ metres.

Properties of a rhombus:

1. The diagonals bisect each other.

2. The diagonals are perpendicular to each other.

3. All sides of a rhombus are equal in length.

Since the diagonals bisect each other, the intersection point O divides each diagonal into two equal segments.

The lengths of the half-diagonals are:

Half of the first diagonal = $\frac{12}{2} = 6$ metres.

Half of the second diagonal = $\frac{16}{2} = 8$ metres.

The distance from the intersection point O to each vertex of the rhombus is the length of the half-diagonal leading to that vertex.

Let the diagonals be AC and BD, intersecting at O. Let AC = 16m and BD = 12m (or vice versa). Then the distances from O to the vertices are $AO = OC = 8$m and $BO = OD = 6$m.

So, the distances from the swing set to the vertices are 6 metres and 8 metres.

However, the options provided do not include 6m or 8m individually, but rather calculations involving the diagonal lengths or half-lengths.

Let's consider a right-angled triangle formed by the intersection of the diagonals. For example, $\triangle AOB$, where O is the intersection. The legs are the half-diagonals, and the hypotenuse is a side of the rhombus (e.g., AB).

Let the half-diagonals be $6$m and $8$m. The angle at O is $90^\circ$. The side length of the rhombus ($s$) is the hypotenuse.

By the Pythagorean theorem:

$s^2 = 6^2 + 8^2$

$s^2 = 36 + 64$

$s^2 = 100$

$s = \sqrt{100} = 10$ metres.

The side length of the rhombus is 10 metres.

Now let's evaluate the given options:

(A) $\sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20$ metres.

(B) $\frac{1}{2} \sqrt{12^2 + 16^2} = \frac{1}{2} \times 20 = 10$ metres.

(C) $\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ metres.

(D) Both B and C are correct.

Options (B) and (C) both calculate the side length of the rhombus, which is 10 metres.

While the actual distances from the swing set (O) to the vertices are 6m and 8m, these are not presented as options. Options (B) and (C) provide a value related to the rhombus geometry, specifically the side length.

Given that options (B) and (C) yield the same result (10m) and option (D) suggests they are both correct, the question likely intends to ask for a distance calculated this way, or is potentially poorly phrased. In the context of the provided options, 10 metres is a significant distance related to the rhombus, derived from the half-diagonals.

Both expressions $\frac{1}{2} \sqrt{12^2 + 16^2}$ and $\sqrt{6^2 + 8^2}$ correctly calculate the side length of the rhombus.

Based on the options, the intended answer relates to the side length calculation.

The correct option is (D) Both B and C are correct.

A quadrilateral is a closed, two-dimensional geometric figure with four straight sides and four vertices.

A quadrilateral has:

Sides: 4

Angles: 4

Vertices: 4

Diagonals: 2

A convex quadrilateral is a quadrilateral where for every side, all other vertices lie on the same side of the line containing that side. Also, all interior angles of a convex quadrilateral are less than $180^\circ$. The two diagonals lie entirely inside the quadrilateral.

A concave quadrilateral (also called a non-convex quadrilateral) is a quadrilateral where at least one of its interior angles is greater than $180^\circ$. In a concave quadrilateral, at least one diagonal lies partly or entirely outside the quadrilateral.

As per the question, you are required to draw a rough sketch for each type. A convex quadrilateral typically looks like a standard square, rectangle, parallelogram, etc., while a concave quadrilateral will have an 'inward-pointing' vertex.

The Angle Sum Property of a quadrilateral states that the sum of the measures of the four interior angles of any quadrilateral is $360^\circ$.

Given the angles of the quadrilateral are $x^\circ, (x+10)^\circ, (x+20)^\circ,$ and $(x+30)^\circ$.

According to the Angle Sum Property, the sum of these angles must be $360^\circ$.

So, we can write the equation:

Combine the terms with $x$ and the constant terms:

$x+x+x+x+10+20+30 = 360$

$4x + 60 = 360$

Subtract 60 from both sides:

$4x = 360 - 60$

$4x = 300$

Divide by 4:

$x = \frac{300}{4}$

$x = 75$

Thus, the value of $x$ is $75$.

We know that the sum of the interior angles of a quadrilateral is $360^\circ$ (Angle Sum Property of a quadrilateral).

Let the measures of the three given angles be $\angle A = 75^\circ$, $\angle B = 90^\circ$, and $\angle C = 105^\circ$. Let the measure of the fourth angle be $\angle D$.

According to the Angle Sum Property:

Substitute the given values into the equation:

$75^\circ + 90^\circ + 105^\circ + \angle D = 360^\circ$

Summing the known angles:

$270^\circ + \angle D = 360^\circ$

Subtract $270^\circ$ from both sides to find $\angle D$:

$\angle D = 360^\circ - 270^\circ$

$\angle D = 90^\circ$

The measure of the fourth angle is $90^\circ$.

The angles of the quadrilateral are $75^\circ, 90^\circ, 105^\circ,$ and $90^\circ$. Since all angles are less than $180^\circ$, this is a convex quadrilateral. As it has two right angles, it could be a trapezoid (specifically, a right trapezoid) or a general quadrilateral with two right angles. Without more information about the side lengths or parallelism of sides, we cannot classify it as a rectangle, square, or parallelogram.

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

Two properties of a parallelogram related to its sides and angles are:

1. Sides: The opposite sides of a parallelogram are equal in length.

2. Angles: The opposite angles of a parallelogram are equal in measure.

Given that ABCD is a parallelogram and $\angle A = 60^\circ$.

In a parallelogram, opposite angles are equal.

Therefore, $\angle C = \angle A$.

In a parallelogram, consecutive angles are supplementary (their sum is $180^\circ$).

Therefore, $\angle A + \angle B = 180^\circ$.

Substitute the value of $\angle A$:

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

Subtract $60^\circ$ from both sides:

Similarly, consecutive angles $\angle B$ and $\angle C$ are supplementary, and $\angle C$ and $\angle D$ are supplementary, and $\angle D$ and $\angle A$ are supplementary. Also, opposite angles $\angle B$ and $\angle D$ are equal.

Therefore,

The measures of the angles are:

$\angle B = 120^\circ$

$\angle C = 60^\circ$

$\angle D = 120^\circ$

Given:

The ratio of adjacent sides of a parallelogram is $3:4$.

The perimeter of the parallelogram is 84 cm.

To Find:

The lengths of the sides of the parallelogram.

Solution:

Let the adjacent sides of the parallelogram be $3x$ and $4x$ cm, where $x$ is a common ratio.

In a parallelogram, opposite sides are equal. So, the four sides are $3x$, $4x$, $3x$, and $4x$ cm.

The perimeter of a parallelogram is the sum of the lengths of its four sides, which is also equal to twice the sum of the lengths of its adjacent sides.

Perimeter $= 2 \times (\text{Length of one adjacent side} + \text{Length of the other adjacent side})$

$84 = 2 \times (3x + 4x)$

$84 = 2 \times (7x)$

$84 = 14x$

To find the value of $x$, divide both sides by 14:

$x = \frac{84}{14}$

$x = 6$

Now, substitute the value of $x$ back into the expressions for the side lengths:

Length of one adjacent side $= 3x = 3 \times 6 = 18$ cm.

Length of the other adjacent side $= 4x = 4 \times 6 = 24$ cm.

Since opposite sides are equal, the lengths of the sides of the parallelogram are 18 cm, 24 cm, 18 cm, and 24 cm.

Two properties of the diagonals of a parallelogram are:

1. The diagonals of a parallelogram bisect each other.

2. The diagonals of a parallelogram divide it into two congruent triangles.

Given that the diagonals of a parallelogram bisect each other at O, and AC is one of the diagonals with a length of 10 cm.

Since the diagonals bisect each other at O, this means that O is the midpoint of AC and BD.

Therefore, the diagonal AC is divided into two equal parts by O, namely AO and OC.

Substitute the given length of AC:

$AO = \frac{1}{2} \times 10$ cm

$AO = 5$ cm

Thus, the length of AO is 5 cm.

A rectangle is a parallelogram in which one of the angles is a right angle ($90^\circ$). Since opposite angles of a parallelogram are equal and consecutive angles are supplementary, if one angle is $90^\circ$, all angles must be $90^\circ$. Thus, a rectangle is a parallelogram with four right angles.

One property of a rectangle that is not true for a general parallelogram, related to its angles, is:

All four interior angles of a rectangle are right angles ($90^\circ$). In a general parallelogram, only opposite angles are equal, and consecutive angles are supplementary; they are not necessarily $90^\circ$ unless the parallelogram is a rectangle or a square.

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

One property of a rhombus that is not true for a general parallelogram, related to its sides, is:

All four sides of a rhombus are equal in length. In a general parallelogram, only opposite sides are equal in length; adjacent sides are not necessarily equal unless the parallelogram is a rhombus or a square.

A square is a quadrilateral with four equal sides and four right angles. It is also defined as a parallelogram with all sides equal and all angles equal to $90^\circ$.

A square is considered a special case of a rectangle because:

A rectangle is defined as a parallelogram with four right angles. A square is a parallelogram with four right angles. Therefore, a square meets the definition of a rectangle. The difference is that a square has the additional property that all its sides are equal, which is not a requirement for a general rectangle.

A square is considered a special case of a rhombus because:

A rhombus is defined as a parallelogram with four equal sides. A square is a parallelogram with four equal sides. Therefore, a square meets the definition of a rhombus. The difference is that a square has the additional property that all its angles are right angles, which is not a requirement for a general rhombus.

In summary, a square possesses all the properties of both a rectangle (four right angles) and a rhombus (four equal sides), in addition to being a parallelogram.

Given:

The lengths of the diagonals of a rectangle are $2x + 8$ and $3x - 5$.

To Find:

The value of $x$ and the length of each diagonal.

Solution:

A key property of a rectangle is that its diagonals are equal in length.

Therefore, we can set the expressions for the lengths of the diagonals equal to each other:

To solve for $x$, rearrange the equation by moving the terms with $x$ to one side and the constant terms to the other side.

Subtract $2x$ from both sides:

$8 = 3x - 2x - 5$

$8 = x - 5$

Add 5 to both sides:

$8 + 5 = x$

$13 = x$

So, the value of $x$ is 13.

Now, find the length of each diagonal by substituting the value of $x=13$ into the expressions:

Length of the first diagonal $= 2x + 8 = 2(13) + 8 = 26 + 8 = 34$.

Length of the second diagonal $= 3x - 5 = 3(13) - 5 = 39 - 5 = 34$.

The lengths are equal, which confirms our value of $x$.

The length of each diagonal is 34.

Given:

The lengths of the diagonals of a rhombus are 6 cm and 8 cm.

To Find:

The length of each side of the rhombus.

Solution:

Let the diagonals of the rhombus be $d_1 = 6$ cm and $d_2 = 8$ cm.

In a rhombus, the diagonals bisect each other at right angles. Let the point of intersection of the diagonals be O.

When the diagonals intersect, they form four right-angled triangles. The sides of the rhombus are the hypotenuses of these right-angled triangles.

The lengths of the segments of the diagonals are half the lengths of the diagonals:

Length of half of the first diagonal $= \frac{d_1}{2} = \frac{6}{2} = 3$ cm.

Length of half of the second diagonal $= \frac{d_2}{2} = \frac{8}{2} = 4$ cm.

Consider one of the right-angled triangles formed by the intersection of the diagonals. The lengths of the two legs of this triangle are 3 cm and 4 cm, and the hypotenuse is the side of the rhombus. Let the side length of the rhombus be $s$.

Using the Pythagorean Theorem ($leg_1^2 + leg_2^2 = hypotenuse^2$) in this right-angled triangle:

$9 + 16 = s^2$

$25 = s^2$

Taking the square root of both sides (and considering only the positive value for length):

$s = \sqrt{25}$

$s = 5$ cm

Since all sides of a rhombus are equal, the length of each side of the rhombus is 5 cm.

A trapezium (or trapezoid in American English) is a quadrilateral with at least one pair of parallel sides. The parallel sides are called bases, and the non-parallel sides are called legs or lateral sides.

A trapezium is called an isosceles trapezium if its non-parallel sides (legs) are equal in length. In an isosceles trapezium, the base angles (angles on the same base) are equal, and the diagonals are equal in length.

A kite is a quadrilateral with two pairs of equal-length sides that are adjacent to each other. This means a kite has exactly two distinct pairs of adjacent sides that are equal in length.

Two properties of a kite related to its diagonals and angles are:

1. Diagonals: The diagonals of a kite are perpendicular to each other. Also, one diagonal bisects the other diagonal (the diagonal connecting the vertices where unequal sides meet is bisected by the other diagonal). The diagonal connecting the vertices where equal sides meet is the axis of symmetry and bisects the angles at these vertices.

2. Angles: One pair of opposite angles (the angles between unequal sides) are equal.

Given:

In $\triangle ABC$, D, E, and F are the midpoints of sides AB, BC, and CA respectively.

AB = 8 cm, BC = 10 cm, CA = 12 cm.

To Find:

The perimeter of $\triangle DEF$.

Theorem Used:

The theorem used is the Midpoint Theorem.

The Midpoint Theorem states that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is half of the third side.

Solution:

Since D and E are the midpoints of sides AB and BC respectively, by the Midpoint Theorem:

Substitute the length of AC:

$DE = \frac{1}{2} \times 12$ cm

$DE = 6$ cm

Since E and F are the midpoints of sides BC and CA respectively, by the Midpoint Theorem:

Substitute the length of AB:

$EF = \frac{1}{2} \times 8$ cm

$EF = 4$ cm

Since F and D are the midpoints of sides CA and AB respectively, by the Midpoint Theorem:

Substitute the length of BC:

$FD = \frac{1}{2} \times 10$ cm

$FD = 5$ cm

The perimeter of $\triangle DEF$ is the sum of the lengths of its sides:

Perimeter of $\triangle DEF = DE + EF + FD$

Perimeter of $\triangle DEF = 6$ cm + 4 cm + 5 cm

Perimeter of $\triangle DEF = 15$ cm

The perimeter of $\triangle DEF$ is 15 cm.

Given:

ABCD is a parallelogram.

E is the midpoint of side AD.

A line through A parallel to EC cuts BC at F.

To Show:

ABFC is a parallelogram.

Proof that F is the midpoint of BC:

In parallelogram ABCD, we know that AD is parallel to BC and AD is equal in length to BC.

Since E is the midpoint of AD,

From (i) and the fact that AD = BC, we have:

Consider the quadrilateral AECF. We are given that the line through A parallel to EC cuts BC at F. Thus, AF is parallel to EC.

Also, since AD || BC, the segment AE (which lies on AD) is parallel to the segment FC (which lies on BC).

Since quadrilateral AECF has both pairs of opposite sides parallel (AF || EC and AE || FC), it is a parallelogram.

In a parallelogram, opposite sides are equal in length. Therefore, in parallelogram AECF, we have:

Combining equation (ii) and (iii):

Since F is a point on the line segment BC and its distance from C is half the length of BC, F must be the midpoint of BC.

Discussion regarding ABFC being a parallelogram:

The question asks to show that ABFC is a parallelogram. Assuming the vertices are listed in consecutive order (A to B, B to F, F to C, C to A), for ABFC to be a parallelogram, it must satisfy one of the parallelogram properties, such as:

1. Opposite sides are parallel: AB || FC and AF || BC.

2. Opposite sides are equal: AB = FC and AF = BC.

3. Diagonals bisect each other: The midpoint of diagonal AF is the same as the midpoint of diagonal BC.

We have established that F is the midpoint of BC. Therefore, the midpoint of the diagonal BC of quadrilateral ABFC is F.

For the diagonals of ABFC (AF and BC) to bisect each other, the midpoint of the diagonal AF must also be F. The midpoint of the line segment AF is the point that divides AF into two equal parts. If the midpoint of AF is F, this would imply that the length of AF is 0, which means points A and F must coincide (A = F).

If A = F, then point A lies on the line BC. Since AD || BC, if A lies on BC, then the entire line AD must coincide with the line BC. This implies that the parallelogram ABCD is degenerate (all vertices lie on a single line).

In a non-degenerate parallelogram ABCD, A does not lie on the line BC, so A is not equal to F.

Let's consider the opposite sides AB and FC. We know AB || DC and AB = DC. We also know FC = $\frac{1}{2}$ BC. In a general parallelogram, there is no relation between AB (or DC) and $\frac{1}{2}$ BC that makes them equal or parallel (unless ABCD is degenerate). Specifically, AB is parallel to DC, not typically to BC (or FC which lies on BC).

Therefore, based on the standard definition of a quadrilateral ABFC (with vertices in consecutive order) and the properties derived from the given information, ABFC is generally not a parallelogram in a non-degenerate case.

It is possible that the question contains a typo and intended to ask to prove that AECF is a parallelogram, or ABFE is a parallelogram, or EFCD is a parallelogram, all of which are true when E and F are midpoints of opposite sides in a parallelogram.

However, strictly following the question as stated, ABFC with vertices A, B, F, C in order is a parallelogram only in a degenerate case where A, B, C, D are collinear.

Yes, a quadrilateral with all sides equal can be a parallelogram.

A quadrilateral with all sides equal is called a rhombus.

A rhombus is defined as a parallelogram in which all four sides are equal in length.

Alternatively, it can be defined as a quadrilateral with all sides equal. It can then be proven that a quadrilateral with all sides equal has opposite sides parallel, and thus is a parallelogram.

Consider a quadrilateral ABCD where AB = BC = CD = DA.

Join diagonal AC.

In $\triangle ABC$ and $\triangle ADC$:

AB = AD (Given)

BC = DC (Given)

AC = AC (Common side)

Therefore, $\triangle ABC \cong \triangle ADC$ by SSS congruence criterion.

By CPCTC, $\angle BAC = \angle DAC$ and $\angle BCA = \angle DCA$.

$\angle BAC$ and $\angle DCA$ are alternate interior angles for lines AB and DC intersected by transversal AC. Since they are equal, AB || DC.

$\angle BCA$ and $\angle DAC$ are alternate interior angles for lines BC and AD intersected by transversal AC. Since they are equal, BC || AD.

Since both pairs of opposite sides are parallel (AB || DC and BC || AD), the quadrilateral ABCD is a parallelogram.

Therefore, a quadrilateral with all sides equal is a parallelogram, and it is specifically called a rhombus.

No, a quadrilateral with only one pair of opposite sides parallel is generally not a parallelogram. It is a trapezium.

A parallelogram is defined as a quadrilateral in which both pairs of opposite sides are parallel.

If a quadrilateral has only one pair of opposite sides parallel, it fits the definition of a trapezium, not a parallelogram.

For example, consider a quadrilateral ABCD where only AB is parallel to DC, but AD is not parallel to BC. This quadrilateral is a trapezium.

To be a parallelogram, the quadrilateral must satisfy the condition that BC is also parallel to AD.

However, there is a special case. If a quadrilateral has one pair of opposite sides parallel and equal in length, then it is a parallelogram.

Let ABCD be a quadrilateral where AB || DC and AB = DC. Draw a diagonal AC.

In $\triangle ABC$ and $\triangle CDA$:

AB = CD (Given)

AC = AC (Common side)

$\angle BAC = \angle DCA$ (Alternate interior angles, since AB || DC)

Therefore, $\triangle ABC \cong \triangle CDA$ by SAS congruence criterion.

By CPCTC, $\angle BCA = \angle DAC$. These are alternate interior angles for lines BC and AD intersected by transversal AC. Since they are equal, BC || AD.

Since both pairs of opposite sides are parallel (AB || DC and BC || AD), ABCD is a parallelogram.

So, while a quadrilateral with *just* one pair of opposite sides parallel is a trapezium, if that single pair of parallel sides is also equal in length, then the quadrilateral must be a parallelogram.

Therefore, simply having one pair of opposite sides parallel is not sufficient to guarantee that a quadrilateral is a parallelogram.

Given:

PQRS is a parallelogram.

$\angle P = 2 \times \angle Q - 30^\circ$.

To Find:

The measure of each angle ($\angle P, \angle Q, \angle R, \angle S$) of the parallelogram.

Solution:

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

So, we have:

We are given that $\angle P$ is $30^\circ$ less than twice $\angle Q$. We can write this relationship as:

Now, substitute the expression for $\angle P$ from equation (i) into the supplementary angle equation:

$(2\angle Q - 30^\circ) + \angle Q = 180^\circ$

Combine the terms with $\angle Q$:

$3\angle Q - 30^\circ = 180^\circ$

Add $30^\circ$ to both sides of the equation:

$3\angle Q = 180^\circ + 30^\circ$

$3\angle Q = 210^\circ$

Divide both sides by 3 to find $\angle Q$:

$\angle Q = \frac{210^\circ}{3}$

$\angle Q = 70^\circ$

Now that we have the value of $\angle Q$, we can find $\angle P$ using equation (i):

$\angle P = 2\angle Q - 30^\circ$

$\angle P = 2(70^\circ) - 30^\circ$

$\angle P = 140^\circ - 30^\circ$

$\angle P = 110^\circ$

In a parallelogram, opposite angles are equal.

So, $\angle R = \angle P$ and $\angle S = \angle Q$.

The measures of the angles of the parallelogram are:

$\angle P = 110^\circ$

$\angle Q = 70^\circ$

$\angle R = 110^\circ$

$\angle S = 70^\circ$

Given:

The ratio of the angles of a quadrilateral is $3:5:9:13$.

To Find:

The measure of each angle of the quadrilateral.

Solution:

Let the angles of the quadrilateral be $3x, 5x, 9x,$ and $13x$, where $x$ is a common ratio.

The sum of the interior angles of a quadrilateral is $360^\circ$ (Angle Sum Property of a quadrilateral).

So, we can write the equation:

Combine the terms with $x$:

$(3+5+9+13)x = 360^\circ$

$30x = 360^\circ$

Divide both sides by 30 to find the value of $x$:

$x = \frac{360^\circ}{30}$

$x = 12^\circ$

Now, find the measure of each angle by substituting the value of $x=12^\circ$:

First angle $= 3x = 3 \times 12^\circ = 36^\circ$.

Second angle $= 5x = 5 \times 12^\circ = 60^\circ$.

Third angle $= 9x = 9 \times 12^\circ = 108^\circ$.

Fourth angle $= 13x = 13 \times 12^\circ = 156^\circ$.

To verify, check the sum of the angles: $36^\circ + 60^\circ + 108^\circ + 156^\circ = 360^\circ$. The sum is correct.

The measures of the angles of the quadrilateral are $36^\circ, 60^\circ, 108^\circ,$ and $156^\circ$.

Given:

ABCD is a quadrilateral whose diagonals AC and BD intersect at point O, such that AO = OC and BO = OD.

To Prove:

Quadrilateral ABCD is a parallelogram.

Proof:

Consider triangles $\triangle AOB$ and $\triangle COD$.

Therefore, $\triangle AOB \cong \triangle COD$ by SAS congruence criterion.

By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), we have:

The angles $\angle BAO$ and $\angle DCO$ are alternate interior angles formed by transversal AC intersecting lines AB and DC. Since these alternate interior angles are equal, it implies that:

Now, consider triangles $\triangle BOC$ and $\triangle DOA$.

Therefore, $\triangle BOC \cong \triangle DOA$ by SAS congruence criterion.

By CPCTC, we have:

The angles $\angle BCO$ and $\angle DAO$ are alternate interior angles formed by transversal AC intersecting lines BC and AD. Since these alternate interior angles are equal, it implies that:

From (iii), we have AB || DC. From (vi), we have BC || AD.

Since both pairs of opposite sides of quadrilateral ABCD are parallel, ABCD is a parallelogram.

Alternatively, we could use the property that if in a quadrilateral, one pair of opposite sides is equal and parallel, it is a parallelogram.

From (i), we have AB = CD. From (iii), we have AB || DC.

Since one pair of opposite sides (AB and DC) is both equal and parallel, quadrilateral ABCD is a parallelogram.

Thus, it is proved that if the diagonals of a quadrilateral bisect each other, it is a parallelogram.

Given:

In $\triangle XYZ$, L is the midpoint of side XY and M is the midpoint of side XZ.

LM = 5 cm.

To Find:

The length of side YZ.

Theorem Used:

The theorem used is the Midpoint Theorem.

The Midpoint Theorem states that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is half the length of the third side.

Solution:

Since L and M are the midpoints of sides XY and XZ respectively, the line segment LM connects the midpoints of two sides of $\triangle XYZ$.

According to the Midpoint Theorem, the line segment LM is parallel to the third side YZ and its length is half the length of YZ.

We are given that LM = 5 cm.

Substitute this value into equation (i):

$5$ cm $= \frac{1}{2} \times YZ$

To find the length of YZ, multiply both sides of the equation by 2:

$YZ = 2 \times 5$ cm

$YZ = 10$ cm

The length of YZ is 10 cm.

Given:

ABCD is a parallelogram.

To Prove:

(a) Opposite sides are equal, i.e., AB = CD and BC = DA.

(b) Opposite angles are equal, i.e., $\angle A = \angle C$ and $\angle B = \angle D$.

Proof:

Consider the parallelogram ABCD. Draw a diagonal AC.

Now consider the two triangles formed, $\triangle ABC$ and $\triangle CDA$.

Since ABCD is a parallelogram, we know that opposite sides are parallel. So, AB || DC and AD || BC.

Consider AB || DC with transversal AC. The alternate interior angles are equal.

AC is a common side to both triangles.

Consider AD || BC with transversal AC. The alternate interior angles are equal.

By ASA (Angle-Side-Angle) congruence criterion, $\triangle ABC$ is congruent to $\triangle CDA$.

Now, by CPCTC (Corresponding Parts of Congruent Triangles are Congruent), we can state the following:

From $\triangle ABC \cong \triangle CDA$:

Equations (ii) and (iii) prove that the opposite sides of a parallelogram are equal.

Also, from $\triangle ABC \cong \triangle CDA$, by CPCTC:

This means $\angle B = \angle D$. This proves one pair of opposite angles is equal.

Now consider the other pair of opposite angles, $\angle A$ and $\angle C$.

Angle A of the parallelogram is the sum of angles $\angle BAC$ and $\angle DAC$.

Angle C of the parallelogram is the sum of angles $\angle DCA$ and $\angle BCA$.

From the congruence $\triangle ABC \cong \triangle CDA$, we know:

Adding these two equalities:

$\angle BAC + \angle DAC = \angle DCA + \angle BCA$

Substituting from (iv) and (v):

Equations (iv) and (vi) prove that the other pair of opposite angles of a parallelogram are equal.

Hence, it is proved that in a parallelogram, opposite sides are equal and opposite angles are equal.

Given:

ABCD is a parallelogram. Its diagonals AC and BD intersect at point O.

To Prove:

The diagonals bisect each other, i.e., AO = OC and BO = OD.

Proof:

Consider the parallelogram ABCD. Since it is a parallelogram, we know that its opposite sides are parallel (AB || DC and AD || BC).

We will consider the triangles formed by the diagonals and the opposite sides, for example, $\triangle AOB$ and $\triangle COD$.

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

Since AB || DC and BD is a transversal, the alternate interior angles are equal:

Since AB || DC and AC is a transversal, the alternate interior angles are equal:

In a parallelogram, opposite sides are equal in length. (This is a property that can be proven using congruence of triangles as shown in a previous question, or taken as a defining property depending on the context and what is already established).

Now, in $\triangle AOB$ and $\triangle COD$, we have:

Therefore, $\triangle AOB \cong \triangle COD$ by ASA (Angle-Side-Angle) congruence criterion.

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

From $\triangle AOB \cong \triangle COD$, we get:

Since AO = OC, the diagonal AC is bisected at O. Since BO = OD, the diagonal BD is bisected at O.

Thus, the diagonals of a parallelogram bisect each other at their point of intersection O.

Hence, it is proved that the diagonals of a parallelogram bisect each other.

Statement of the Midpoint Theorem:

The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is half of the third side.

Proof:

Given:

$\triangle ABC$ is a triangle.

D is the midpoint of side AB.

E is the midpoint of side AC.

To Prove:

1. DE || BC

2. $DE = \frac{1}{2} BC$

Construction:

Extend the line segment DE to a point F such that DE = EF. Join F to C.

Proof:

Consider $\triangle ADE$ and $\triangle CFE$.

Therefore, $\triangle ADE \cong \triangle CFE$ by SAS congruence criterion.

By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), we have:

From (iii), $\angle ADE$ and $\angle CFE$ are alternate interior angles for lines AB and CF intersected by transversal DF. Since these alternate interior angles are equal, it implies that AD || CF.

Since D lies on AB, AD is part of line AB. So, AB || CF.

Since D is the midpoint of AB, we have AD = DB.

From (ii), AD = CF. By transitivity, DB = CF.

Also, since AD || CF, and D lies on AB, we have DB || CF.

Consider quadrilateral DBCF. From (v) and (vi), we have DB = CF and DB || CF. If one pair of opposite sides of a quadrilateral is equal and parallel, then it is a parallelogram.

Therefore, quadrilateral DBCF is a parallelogram.

In a parallelogram, opposite sides are parallel and equal.

So, in parallelogram DBCF:

Since DE is a part of the line segment DF, it means DE is also parallel to BC.

Also, in parallelogram DBCF:

By construction, DF = DE + EF, and we constructed EF = DE. So, DF = DE + DE = 2DE.

Substitute this into the equality DF = BC:

$2DE = BC$

Dividing by 2, we get:

Thus, it is proved that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of the third side.

Given:

ABCD is a rhombus. Its diagonals AC and BD intersect at point O.

To Show:

1. The diagonals bisect each other (AO = OC and BO = OD).

2. The diagonals are perpendicular to each other (e.g., $\angle AOB = 90^\circ$).

Proof:

Since a rhombus is a special type of parallelogram, it has all the properties of a parallelogram. One key property of a parallelogram is that its diagonals bisect each other.

So, in rhombus ABCD, the diagonals AC and BD bisect each other at O.

This proves the first part: the diagonals bisect each other.

Now, we need to show that the diagonals are perpendicular.

Consider adjacent triangles formed at the intersection of the diagonals, for example, $\triangle AOB$ and $\triangle COB$.

In $\triangle AOB$ and $\triangle COB$:

Therefore, $\triangle AOB \cong \triangle COB$ by SSS (Side-Side-Side) congruence criterion.

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

From $\triangle AOB \cong \triangle COB$, we get:

Angles $\angle AOB$ and $\angle COB$ form a linear pair on the straight line AC. The sum of angles in a linear pair is $180^\circ$.

Since $\angle AOB = \angle COB$, we can substitute $\angle AOB$ for $\angle COB$ in the equation:

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

$2\angle AOB = 180^\circ$

Divide by 2:

Since $\angle AOB = 90^\circ$, the diagonals AC and BD are perpendicular to each other at their intersection point O.

Combining the facts that the diagonals bisect each other (AO = OC, BO = OD) and are perpendicular to each other ($\angle AOB = 90^\circ$), we conclude that the diagonals of a rhombus are perpendicular bisectors of each other.

Thus, it is shown that the diagonals of a rhombus are perpendicular bisectors of each other.

Given:

ABCD is a parallelogram with diagonals AC and BD such that AC = BD.

To Prove:

ABCD is a rectangle (i.e., one of its interior angles is $90^\circ$).

Proof:

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

In a parallelogram, opposite sides are equal.

BC is a common side to both triangles.

We are given that the diagonals are equal.

Therefore, $\triangle ABC \cong \triangle DCB$ by SSS (Side-Side-Side) congruence criterion.

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

From $\triangle ABC \cong \triangle DCB$, we get:

In a parallelogram, consecutive angles are supplementary. So, the sum of adjacent angles is $180^\circ$.

Since $\angle ABC = \angle DCB$, we can substitute $\angle ABC$ for $\angle DCB$ in the equation:

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

$2\angle ABC = 180^\circ$

Divide by 2:

Since $\angle ABC$ is one of the interior angles of the parallelogram ABCD, and we have shown that it is a right angle ($90^\circ$), this proves that the parallelogram ABCD is a rectangle.

(If one angle of a parallelogram is $90^\circ$, all angles are $90^\circ$ because opposite angles are equal and consecutive angles are supplementary).

Thus, it is proved that if the diagonals of a parallelogram are equal, then it is a rectangle.

Given:

ABCD is a quadrilateral.

P, Q, R, and S are the midpoints of sides AB, BC, CD, and DA respectively.

To Show:

PQRS is a parallelogram.

Proof:

Join the diagonal AC of the quadrilateral ABCD.

Consider $\triangle ABC$. P is the midpoint of AB and Q is the midpoint of BC.

By the Midpoint Theorem, the line segment PQ joining the midpoints of sides AB and BC is parallel to the third side AC and is half the length of AC.

Now consider $\triangle ADC$. R is the midpoint of CD and S is the midpoint of DA.

By the Midpoint Theorem, the line segment SR joining the midpoints of sides CD and DA is parallel to the third side AC and is half the length of AC.

From (i) and (ii), we have $PQ = \frac{1}{2} AC$ and $SR = \frac{1}{2} AC$.

Also, from the parallelisms we found:

PQ || AC (from $\triangle ABC$)

SR || AC (from $\triangle ADC$)

If two lines are parallel to the same line, they are parallel to each other.

Consider quadrilateral PQRS. From (iii), we have PQ = SR, and from (iv), we have PQ || SR.

We have shown that one pair of opposite sides of quadrilateral PQRS (namely PQ and SR) is equal in length and parallel.

A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram.

Therefore, quadrilateral PQRS is a parallelogram.

Thus, it is shown that the figure formed by joining the midpoints of the sides of a quadrilateral is a parallelogram.

Given:

ABCD is a square. Its diagonals are AC and BD, intersecting at O.

To Show:

1. AC = BD (Diagonals are equal).

2. The diagonals bisect each other (AO = OC and BO = OD).

3. The diagonals intersect at right angles ($\angle AOB = 90^\circ$).

Proof:

A square is a special type of quadrilateral that is also a rectangle and a rhombus.

Properties of a square:

• All sides are equal (AB = BC = CD = DA).
• All interior angles are $90^\circ$ ($\angle A = \angle B = \angle C = \angle D = 90^\circ$).
• It is a parallelogram (opposite sides are parallel).

Proof that diagonals are equal:

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

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

By CPCTC, the corresponding sides are equal:

Proof that diagonals bisect each other:

Since a square is a parallelogram, its diagonals bisect each other.

Let the diagonals AC and BD intersect at O.

This proves the second part: the diagonals bisect each other.

Proof that diagonals intersect at right angles:

Since a square is a rhombus (all sides equal), and the diagonals of a rhombus are perpendicular bisectors of each other, the diagonals of a square must also be perpendicular bisectors of each other.

Consider $\triangle AOB$ and $\triangle COB$.

Therefore, $\triangle AOB \cong \triangle COB$ by SSS congruence criterion.

By CPCTC, $\angle AOB = \angle COB$.

Since $\angle AOB$ and $\angle COB$ form a linear pair and are equal, they must both be $90^\circ$.

$\angle AOB + \angle COB = 180^\circ$

$2\angle AOB = 180^\circ$

This means the diagonals intersect at right angles.

Combining the results, the diagonals of a square are equal, they bisect each other, and they intersect at right angles. Therefore, the diagonals of a square are perpendicular bisectors of each other.

Thus, it is shown that the diagonals of a square are equal and bisect each other at right angles.

If the diagonals of a quadrilateral are equal and bisect each other at right angles, the quadrilateral is a square.

Given:

ABCD is a quadrilateral. Its diagonals AC and BD intersect at O such that:

1. AC = BD (Diagonals are equal)

2. AO = OC and BO = OD (Diagonals bisect each other)

3. $\angle AOB = \angle BOC = \angle COD = \angle DOA = 90^\circ$ (Diagonals bisect each other at right angles)

To Prove:

ABCD is a square.

Proof:

Since the diagonals bisect each other (AO = OC and BO = OD), quadrilateral ABCD is a parallelogram.

Now, let's use the property that the diagonals are equal (AC = BD).

We have already shown in a previous proof (Question 5) that if the diagonals of a parallelogram are equal, then it is a rectangle.

Since ABCD is a parallelogram and AC = BD, ABCD is a rectangle.

A rectangle has all four angles equal to $90^\circ$.

Now, let's use the property that the diagonals bisect each other at right angles ($\angle AOB = 90^\circ$).

Consider $\triangle AOB$ and $\triangle COB$.

Therefore, $\triangle AOB \cong \triangle COB$ by SAS congruence criterion.

By CPCTC, the corresponding sides are equal:

Since AB and CB are adjacent sides of the parallelogram ABCD, and they are equal, the parallelogram has all its sides equal.

A parallelogram with all sides equal is a rhombus.

So, we have shown that ABCD is both a rectangle and a rhombus.

A quadrilateral that is both a rectangle (all angles are $90^\circ$) and a rhombus (all sides are equal) is a square.

Alternatively, we can combine the properties:

1. Diagonals bisect each other $\implies$ ABCD is a parallelogram.

2. Diagonals are equal $\implies$ Parallelogram is a rectangle (all angles $90^\circ$).

3. Diagonals are perpendicular $\implies$ Parallelogram is a rhombus (all sides equal).

Since ABCD is both a rectangle and a rhombus, it must be a square.

Thus, if the diagonals of a quadrilateral are equal and bisect each other at right angles, it is a square.

Given:

ABCD is a parallelogram.

AP is a perpendicular from A to diagonal BD, so $\angle APB = 90^\circ$.

CQ is a perpendicular from C to diagonal BD, so $\angle CQD = 90^\circ$.

To Show:

$\triangle APB \cong \triangle CQD$.

Proof:

Consider the two triangles $\triangle APB$ and $\triangle CQD$.

Since ABCD is a parallelogram, its opposite sides are equal in length.

We are given that AP $\perp$ BD and CQ $\perp$ BD.

Thus, $\angle APB = \angle CQD$ (both are $90^\circ$).

Since ABCD is a parallelogram, AB is parallel to DC. BD is a transversal intersecting the parallel lines AB and DC.

Therefore, the alternate interior angles formed by the transversal BD are equal.

Now, consider $\triangle APB$ and $\triangle CQD$. We have found:

By AAS (Angle-Angle-Side) congruence criterion, $\triangle APB$ is congruent to $\triangle CQD$. (Note: AB is the side opposite to $\angle APB$ in $\triangle APB$, and CD is the side opposite to $\angle CQD$ in $\triangle CQD$.)

Thus, it is shown that $\triangle APB \cong \triangle CQD$.

Given:

ABCD is a parallelogram. Its diagonals AC and BD intersect at O.

P and Q are points on diagonal BD such that DP = BQ.

To Show:

APCQ is a parallelogram.

Proof:

In a parallelogram, the diagonals bisect each other. So, the diagonals AC and BD bisect each other at O.

We are given that DP = BQ.

Consider the diagonal BD. We have BO = OD.

Since BO = OD, O is the midpoint of BD.

We are given DP = BQ.

Let's look at the segments BP and DQ.

BP = BO - PO (or PO - BO depending on arrangement of P relative to O)

DQ = DO - QO (or QO - DO depending on arrangement of Q relative to O)

Alternatively, we can work with the entire diagonal BD.

Since BO = OD and O is the midpoint of BD, we can write:

BO = OD = $\frac{1}{2} BD$.

We are given DP = BQ.

Consider the segment OP. P lies on BD.

OP = OD - DP (assuming P is between O and D)

OP = OB - BP (assuming P is between B and O)

Consider the segment OQ. Q lies on BD.

OQ = OB - BQ (assuming Q is between O and B)

OQ = OD - DQ (assuming Q is between D and O)

Let's consider the case where P and Q are placed such that B-Q-O-P-D or B-P-O-Q-D. Given DP = BQ, this implies symmetry around O.

Start from the center O. The distance from O to D is OD. The distance from O to B is OB. We know OD = OB.

Point P is on BD such that DP = BQ.

Distance from O to P is OP.

Case 1: P is between O and D, and Q is between O and B.

OP = OD - DP

OQ = OB - BQ

Since OD = OB and DP = BQ, it follows that OD - DP = OB - BQ.

So, OP = OQ.

Case 2: P is between B and O, and Q is between D and O.

OP = OB - PB

OQ = OD - QD

Given DP = BQ.

Since BO = OD and DP = BQ, we have $OD - DP = OB - BQ$ which is equivalent to OP = OQ.

Regardless of the exact position of P and Q (as long as they are on the diagonal BD), if DP = BQ and O is the midpoint of BD, then O is also the midpoint of PQ.

Let's formalize this.

O is the midpoint of BD $\implies$ Vector $\vec{O} = \frac{\vec{B} + \vec{D}}{2}$. (Using position vectors)

We are given $\vec{DP} = \vec{BQ}$. However, vector equality requires both magnitude and direction. Since P and Q are on the line segment BD, the direction is along BD.

Magnitude: $|DP| = |BQ|$. Let's denote the length DP = BQ = k.

Since P lies on BD and DP = k, the position of P relative to D can be written. $P$ is on the line segment BD. There are two possibilities: P is between B and D, or D is between B and P (this would mean P is outside the segment). Assuming P and Q are within the segment BD.

Since O is the midpoint of BD, BO = OD.

Given DP = BQ.

Let's consider distances from O.

OP = |OD - DP| if P is between O and D.

OP = |OB - BP| if P is between B and O.

OQ = |OB - BQ| if Q is between O and B.

OQ = |OD - DQ| if Q is between D and O.

We know OD = OB.

OD = OP + PD $\implies$ OP = OD - PD = OD - DP

OB = OQ + QB $\implies$ OQ = OB - QB = OB - BQ

Since OD = OB and DP = BQ, we have OP = OQ.

So, the diagonal PQ of quadrilateral APCQ is bisected at O.

We already know that the diagonal AC of quadrilateral APCQ is bisected at O (AO = OC).

Thus, the diagonals AC and PQ of quadrilateral APCQ bisect each other at point O.

A quadrilateral whose diagonals bisect each other is a parallelogram.

Therefore, quadrilateral APCQ is a parallelogram.

Thus, it is shown that APCQ is a parallelogram.

Given:

The angles of a quadrilateral are $x^\circ, (x-10)^\circ, (x+30)^\circ,$ and $(2x)^\circ$.

To Find:

The measure of each angle.

The type of quadrilateral.

Whether it can be a parallelogram.

Solution:

The sum of the interior angles of a quadrilateral is $360^\circ$ (Angle Sum Property of a quadrilateral).

So, we can write the equation:

Combine the terms with $x$ and the constant terms:

$x+x+x+2x - 10 + 30 = 360$

$5x + 20 = 360$

Subtract 20 from both sides:

$5x = 360 - 20$

$5x = 340$

Divide by 5:

$x = \frac{340}{5}$

$x = 68$

Now, find the measure of each angle by substituting the value of $x=68^\circ$:

First angle $= x^\circ = 68^\circ$.

Second angle $= (x-10)^\circ = (68-10)^\circ = 58^\circ$.

Third angle $= (x+30)^\circ = (68+30)^\circ = 98^\circ$.

Fourth angle $= (2x)^\circ = (2 \times 68)^\circ = 136^\circ$.

To verify, check the sum of the angles: $68^\circ + 58^\circ + 98^\circ + 136^\circ = 360^\circ$. The sum is correct.

The measures of the angles are $68^\circ, 58^\circ, 98^\circ,$ and $136^\circ$.

What type of quadrilateral could this be?

Since all angles are less than $180^\circ$, this is a convex quadrilateral. Without further information about side lengths or parallelism, it is just a general convex quadrilateral.

Can it be a parallelogram?

In a parallelogram, opposite angles are equal. Let's check if any pair of opposite angles are equal among the calculated angles ($68^\circ, 58^\circ, 98^\circ, 136^\circ$).

The angles are $68^\circ, 58^\circ, 98^\circ, 136^\circ$. None of these values are repeated, so there are no pairs of equal opposite angles.

Also, in a parallelogram, consecutive angles are supplementary (sum up to $180^\circ$). Let's check the sums of adjacent pairs:

$68^\circ + 58^\circ = 126^\circ \neq 180^\circ$

$58^\circ + 98^\circ = 156^\circ \neq 180^\circ$

$98^\circ + 136^\circ = 234^\circ \neq 180^\circ$

$136^\circ + 68^\circ = 204^\circ \neq 180^\circ$

Since neither the opposite angles are equal nor the consecutive angles are supplementary, this quadrilateral cannot be a parallelogram.

Hierarchy of Quadrilaterals:

The hierarchy of quadrilaterals shows the relationships between different types of quadrilaterals, moving from the most general to the most specific. A more specific type of quadrilateral inherits all the properties of the more general types above it in the hierarchy.

Here is a common way to represent the hierarchy:

Quadrilateral

$\downarrow$

Trapezium / Trapezoid (At least one pair of parallel sides)

Kite (Two pairs of equal adjacent sides)

$\downarrow$ (Isosceles Trapezium is a special Trapezium)

$\downarrow$ (Rhombus is a special Kite)

Parallelogram (Both pairs of opposite sides are parallel)

$\swarrow$ $\searrow$

Rhombus (Parallelogram with all sides equal)

Rectangle (Parallelogram with all angles $90^\circ$)

$\downarrow$

$\downarrow$

$\searrow$ $\swarrow$

Square (Rectangle with all sides equal, or Rhombus with all angles $90^\circ$)

Explanation of Inherited Properties:

Moving down the hierarchy means adding more specific properties. A figure lower down has all the properties of the figures above it from which it descends, plus some additional properties.

1. From Quadrilateral to Parallelogram:

A quadrilateral only has 4 sides and 4 angles (summing to $360^\circ$).

A parallelogram inherits these basic properties and adds:

• Opposite sides are parallel.
• Opposite sides are equal in length.
• Opposite angles are equal.
• Consecutive angles are supplementary.
• Diagonals bisect each other.

2. From Parallelogram to Rhombus or Rectangle:

Both Rhombus and Rectangle inherit all the properties of a parallelogram.

A Rhombus adds the property:

• All four sides are equal.
• Diagonals are perpendicular bisectors of each other.
• Diagonals bisect the angles of the rhombus.

A Rectangle adds the property:

• All four angles are right angles ($90^\circ$).
• Diagonals are equal in length and bisect each other. (The "bisect each other" part is inherited from parallelogram, but the "equal" part is new).

3. From Rhombus and Rectangle to Square:

A square inherits all the properties of both a rhombus and a rectangle.

A Square combines the properties of a rhombus and a rectangle:

• All four sides are equal (from Rhombus).
• All four angles are right angles ($90^\circ$) (from Rectangle).
• Opposite sides are parallel (inherited from Parallelogram).
• Opposite angles are equal (inherited from Parallelogram).
• Consecutive angles are supplementary (inherited from Parallelogram).
• Diagonals are equal in length (from Rectangle).
• Diagonals bisect each other (from Parallelogram/Rhombus/Rectangle).
• Diagonals are perpendicular bisectors of each other (from Rhombus).
• Diagonals bisect the angles (which are $90^\circ$), so diagonals bisect the angles into $45^\circ$ angles (from Rhombus).

So, as we move down the hierarchy, the quadrilaterals become more specific and possess a larger set of properties.