Chapter 4 Quadrilaterals (Class 8 - Latest Maths NCERT (Ganita Prakash I) Solutions)

Case (i): Rectangle ABCD

Given: $ABCD$ is a rectangle where diagonals $AC$ and $BD$ intersect at $O$. $\angle CAB = 30^\circ$.

To Find: All internal angles of the triangles formed by the diagonals.

Solution:

In a rectangle, diagonals are equal and bisect each other.

In $\triangle OAB$, since $OA = OB$, it is an isosceles triangle.

By the angle sum property in $\triangle OAB$:

Since $\angle DAB = 90^\circ$ (Vertex angle of a rectangle):

$\angle DAC = 90^\circ - \angle CAB = 90^\circ - 30^\circ = 60^\circ$

In $\triangle OAD$, since $OA = OD$:

$\angle ODA = \angle OAD = 60^\circ$

$\angle AOD = 180^\circ - (60^\circ + 60^\circ) = 60^\circ$

By symmetry and property of alternate interior angles ($AB \parallel DC$):

$\angle ODC = 30^\circ$, $\angle OCD = 30^\circ$, $\angle OCB = 60^\circ$, $\angle OBC = 60^\circ$.

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Case (ii): Rectangle PQRS

Given: $PQRS$ is a rectangle where diagonals intersect at $O$ and $\angle QOR = 110^\circ$.

To Find: All other internal angles.

Solution:

In $\triangle QOR$, $OQ = OR$. Let $\angle OQR = \angle ORQ = x$.

$2x = 70^\circ \implies x = 35^\circ$

So, $\angle OQR = 35^\circ$ and $\angle ORQ = 35^\circ$.

Since $\angle PQR = 90^\circ$:

By symmetry, the angles at the vertices are $35^\circ$ and $55^\circ$. The remaining intersection angle is:

$\angle POQ = 180^\circ - 110^\circ = 70^\circ$ (Linear pair)

To Find: The type of quadrilateral formed in each case.

Solution:

Based on the geometric properties of quadrilaterals, if the diagonals are equal and bisect each other, the figure is always a Rectangle. If they are also perpendicular, it becomes a Square.

Angle between Diagonals	Type of Quadrilateral	Reason
$30^\circ$	Rectangle	Equal diagonals bisecting each other.
$40^\circ$	Rectangle	Equal diagonals bisecting each other.
$90^\circ$	Square	Equal diagonals bisecting perpendicularly.
$140^\circ$	Rectangle	Equal diagonals bisecting each other.

Note: For (iv), $140^\circ$ is the obtuse angle of the intersection. The acute angle would be $180^\circ - 140^\circ = 40^\circ$. Both represent the same rectangle as in case (ii).

Given: $PL$ and $AM$ are two diameters of a circle with centre $O$, and $PL \perp AM$.

To Find: The shape of the quadrilateral $APML$.

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Proof:

In quadrilateral $APML$, the diagonals are $PL$ and $AM$. We analyze their properties:

1. Since both $PL$ and $AM$ are diameters of the same circle:

2. Since $O$ is the centre of the circle:

This implies that the diagonals $PL$ and $AM$ bisect each other at $O$.

3. It is given that the diameters are perpendicular:

Conclusion:

A quadrilateral whose diagonals are equal, bisect each other, and intersect at $90^\circ$ is a Square.

Thus, the figure $APML$ is a Square.

Given: Two sticks of equal length and a piece of thread.

Construction Required: To construct an angle of $90^\circ$.

Solution:

We can use the geometric property that a quadrilateral whose diagonals are equal and bisect each other is a Rectangle, and every vertex angle of a rectangle is $90^\circ$.

1. Take the two sticks of equal length. Use the thread to find the exact midpoint of each stick by folding the measured thread in half.

2. Place one stick over the other so they cross each other. Tie them together firmly at their midpoints using the thread.

3. Now, use the remaining thread to connect the four endpoints of the sticks to form a boundary.

4. The resulting figure is a Rectangle because its diagonals (the sticks) are equal in length and bisect each other (tied at midpoints).

By following this method, the four corners where the thread meets the ends of the sticks will form exact $90^\circ$ angles.

To Find: Whether the property "opposite sides are parallel and equal" is sufficient to define a rectangle.

Solution:

No, the property that opposite sides are parallel and equal cannot be chosen as the sole definition of a rectangle.

A quadrilateral with opposite sides parallel and equal is the definition of a Parallelogram.

While it is true that every rectangle is a parallelogram, the converse is not true. In a general parallelogram, the internal angles do not have to be $90^\circ$. They can be any pair of supplementary angles (e.g., $60^\circ$ and $120^\circ$).

To specifically define a Rectangle, we need one additional condition added to the parallelogram property:

1. At least one angle is $90^\circ$ (which forces all others to be $90^\circ$ in a parallelogram).

OR

2. The diagonals must be equal in length.

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Conclusion:

Every quadrilateral with opposite sides parallel and equal is a parallelogram, but it is not necessarily a rectangle. Therefore, this description is insufficient to serve as a definition for a rectangle.

(i) Parallelogram PEAR

Given: $PEAR$ is a parallelogram with $\angle P = 40^\circ$.

To Find: Remaining angles $\angle E, \angle A, \text{ and } \angle R$.

Solution:

In a parallelogram, opposite angles are equal and adjacent angles are supplementary.

$\angle E = 180^\circ - 40^\circ = 140^\circ$

Remaining angles of $PEAR$ are $140^\circ, 40^\circ, \text{ and } 140^\circ$.

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(ii) Parallelogram PQRS

Given: $PQRS$ is a parallelogram with $\angle P = 110^\circ$.

To Find: Remaining angles $\angle Q, \angle R, \text{ and } \angle S$.

Solution:

Remaining angles of $PQRS$ are $70^\circ, 110^\circ, \text{ and } 70^\circ$.

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(iii) Rhombus XUVW

Given: $XUVW$ is a rhombus. Side marks indicate all sides are equal. Diagonal $XV$ is drawn and $\angle UVX = 30^\circ$.

To Find: All internal angles of the rhombus.

Solution:

In a rhombus, diagonals bisect the vertex angles.

Total $\angle V = \angle WVX + \angle UVX = 30^\circ + 30^\circ = 60^\circ$

The angles of the rhombus are $60^\circ, 120^\circ, 60^\circ, \text{ and } 120^\circ$.

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(iv) Rhombus OAIE

Given: $OAIE$ is a rhombus. Diagonal $OE$ is drawn and $\angle AEO = 20^\circ$.

To Find: All internal angles of the rhombus.

Solution:

Diagonal $OE$ bisects $\angle E$.

$\angle IEO = \angle AEO = 20^\circ$

Total vertex angle $\angle E = 20^\circ + 20^\circ = 40^\circ$

The angles of the rhombus are $140^\circ, 40^\circ, 140^\circ, \text{ and } 40^\circ$.

To Construct: A parallelogram with diagonals $d_1 = 7 \text{ cm}$, $d_2 = 5 \text{ cm}$, and intersection angle $140^\circ$.

Property: The diagonals of a parallelogram bisect each other.

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Construction Steps:

1. Draw a line segment $AC = 7 \text{ cm}$ (the first diagonal). Mark its midpoint as $O$.

2. At point $O$, use a protractor to draw a line $XY$ making an angle of $140^\circ$ with $OA$.

3. Since the second diagonal is $5 \text{ cm}$, set the compass to $2.5 \text{ cm}$ ($\frac{5}{2} = 2.5$).

4. Placing the compass at $O$, cut arcs on both sides of the line $XY$. Mark these points as $B$ and $D$.

5. Join $AB, BC, CD, \text{ and } DA$.

Conclusion: $ABCD$ is the required parallelogram.

To Construct: A rhombus with diagonals $d_1 = 4 \text{ cm}$ and $d_2 = 5 \text{ cm}$.

Property: The diagonals of a rhombus bisect each other at right angles ($90^\circ$).

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Construction Steps:

1. Draw a line segment $AC = 5 \text{ cm}$.

2. Draw the perpendicular bisector of $AC$. Let it intersect $AC$ at point $O$.

3. The other diagonal is $4 \text{ cm}$. Set the compass to $2 \text{ cm}$ ($\frac{4}{2} = 2$).

4. From $O$, cut arcs of $2 \text{ cm}$ on both sides of the perpendicular bisector. Mark these points as $B$ and $D$.

5. Join $AB, BC, CD, \text{ and } DA$.

Conclusion: $ABCD$ is the required rhombus.

Given: Two equilateral triangles, each with a side length of $4 \text{ cm}$, are joined together along one of their sides.

To Find: The sides and angles of the resulting quadrilateral.

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Solution:

In an equilateral triangle, all three sides are equal and all three internal angles are $60^\circ$.

When two such triangles are joined along a common side, that common side becomes a diagonal of the resulting quadrilateral.

1. Sides: Since each triangle has all sides equal to $4 \text{ cm}$, the four outer sides of the resulting quadrilateral will also be equal.

Since all four sides are equal, the resulting quadrilateral is a Rhombus.

2. Angles: Each angle in an equilateral triangle is $60^\circ$.

Two opposite vertices of the quadrilateral will consist of a single $60^\circ$ angle from each triangle.

The other two opposite vertices are formed by joining two angles of the triangles together.

Conclusion: The resulting quadrilateral is a Rhombus with all sides equal to $4 \text{ cm}$ and internal angles of $60^\circ, 120^\circ, 60^\circ, \text{ and } 120^\circ$.

Given: Diagonals of the kite are $d_1 = 6 \text{ cm}$ and $d_2 = 8 \text{ cm}$.

Property: In a kite, diagonals intersect at right angles ($90^\circ$), and one diagonal (the main diagonal) bisects the other diagonal (the cross diagonal).

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Construction Steps:

1. Draw the shorter diagonal $AC = 6 \text{ cm}$.

2. Draw the perpendicular bisector of $AC$. Let it meet $AC$ at point $O$.

3. The longer diagonal is $8 \text{ cm}$. For a kite, the intersection point $O$ does not need to be the midpoint of the longer diagonal. However, the total length must be $8 \text{ cm}$.

4. From $O$, mark a point $B$ at a distance of $3 \text{ cm}$ on one side of the bisector and a point $D$ at a distance of $5 \text{ cm}$ on the other side (Total $3 + 5 = 8 \text{ cm}$).

5. Join $AB, BC, CD, \text{ and } DA$.

Conclusion: $ABCD$ is the required Kite.

Property: In a trapezium, the angles on the same side of the non-parallel legs (consecutive interior angles) are supplementary, i.e., their sum is $180^\circ$.

(i) First Trapezium:

The top and bottom sides are parallel (indicated by arrows).

1. Angle at bottom left is $135^\circ$. Let the top left angle be $x$.

$x + 135^\circ = 180^\circ$

$x = 180^\circ - 135^\circ = 45^\circ$

2. Angle at bottom right is $105^\circ$. Let the top right angle be $y$.

$y + 105^\circ = 180^\circ$

$y = 180^\circ - 105^\circ = 75^\circ$

Result: The remaining angles are $45^\circ$ and $75^\circ$.

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(ii) Second Trapezium:

This is an Isosceles Trapezium as indicated by the equal side markings on the non-parallel sides.

1. In an isosceles trapezium, base angles are equal and top angles are equal.

2. Given top left angle is $100^\circ$. Thus, the top right angle is also $100^\circ$.

3. Let the bottom angles be $z$. Since they are supplementary to the top angles:

$z + 100^\circ = 180^\circ$

Result: The remaining angles are $100^\circ, 80^\circ, \text{ and } 80^\circ$.

Venn Diagram Description:

1. The largest sets are Parallelograms and Kites. They intersect.

2. The intersection of Parallelograms and Kites is the set of Rhombuses.

3. Rectangles are a subset of Parallelograms.

4. The intersection of Rhombuses and Rectangles is the set of Squares. A Square is also a subset of Kites.

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Answers:

(i) A quadrilateral that is both a kite (adjacent sides equal) and a parallelogram (opposite sides equal and parallel) must have all four sides equal. Therefore, it is a Rhombus.

(ii) Yes. A Square is both a kite (it has adjacent equal sides) and a rectangle (it has opposite equal sides and all angles are $90^\circ$).

(iii) No, every kite is not a rhombus. In a kite, only two pairs of adjacent sides are equal. In a rhombus, all four sides must be equal.

Correct Relationship: Every Rhombus is a Kite, but every Kite is not necessarily a Rhombus. Rhombus is a special case of a kite where all sides are equal.

Given: $PAIR$ and $RODS$ are two rectangles. In rectangle $PAIR$, $PR = 5 \text{ cm}$. In rectangle $RODS$, $RS = 5 \text{ cm}$. The angle $\angle ORI = 30^\circ$. Point $O$ lies on the side $AI$ of rectangle $PAIR$.

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

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Solution:

In rectangle $PAIR$, the side $RI$ is perpendicular to side $AI$.

$\angle RIA = 90^\circ$ (Vertex angle of a rectangle)

Since point $O$ lies on the segment $AI$, we consider the right-angled triangle $ROI$.

In $\triangle ROI$:

$\angle ORI = 30^\circ$ (Given)

$\angle RIO = 90^\circ$

Using the angle sum property of a triangle:

$\angle ROI = 180^\circ - (90^\circ + 30^\circ)$

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

Now, consider the rectangle $RODS$. In a rectangle, the angle at each vertex is $90^\circ$.

$\angle ROD = 90^\circ$ (Vertex angle of rectangle $RODS$)

From the figure, $\angle ROD$ is composed of $\angle ROI$ and $\angle IOD$.

$\angle IOD = \angle ROD - \angle ROI$

$\angle IOD = 90^\circ - 60^\circ$

$\angle IOD = 30^\circ$

Conclusion: The measure of $\angle IOD$ is $30^\circ$.

Property: In a square, the diagonals are equal in length and bisect each other at right angles ($90^\circ$).

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Construction Steps:

1. Draw a line segment $AC = 6 \text{ cm}$. This will be the first diagonal of the square.

2. Using a compass, construct the perpendicular bisector of $AC$. To do this, draw arcs from $A$ and $C$ with a radius more than half of $AC$. Let the bisector intersect $AC$ at point $O$.

3. Point $O$ is the midpoint of $AC$, so $OA = OC = 3 \text{ cm}$.

4. Since the diagonals of a square are equal ($6 \text{ cm}$) and bisected, the other diagonal must also have segments of $3 \text{ cm}$ from the center $O$.

5. With $O$ as center and radius $3 \text{ cm}$, draw arcs on both sides of the perpendicular bisector to mark points $B$ and $D$.

6. Join $AB, BC, CD, \text{ and } DA$.

Conclusion: $ABCD$ is the required Square with diagonal length $6 \text{ cm}$.

Geometric Reasoning:

Let the side of the square $CASE$ be $s$. Since $U, V, W, \text{ and } X$ are midpoints, they divide each side into two equal segments of length $\frac{s}{2}$.

1. Consider the four corner triangles: $\triangle CVU, \triangle AUX, \triangle SXW, \text{ and } EWV$.

2. In each triangle, the two legs are equal to $\frac{s}{2}$ and the included angle is $90^\circ$ (vertex of the square $CASE$).

3. By SAS Congruence Criterion, all four triangles are congruent.

4. Therefore, their hypotenuses are equal: $UV = UX = XW = WV$. This makes $UVWX$ a Rhombus.

5. In $\triangle CVU$ (an isosceles right triangle), $\angle CUV = \angle CVU = 45^\circ$. Similarly, in $\triangle AUX$, $\angle AUX = 45^\circ$.

6. At point $U$ on the straight line $CA$:

$\angle CUV + \angle VUX + \angle AUX = 180^\circ$ (Angles on a straight line)

$45^\circ + \angle VUX + 45^\circ = 180^\circ$

$\angle VUX = 180^\circ - 90^\circ = 90^\circ$

7. A rhombus with an internal angle of $90^\circ$ is a Square.

Conclusion: The quadrilateral $UVWX$ is a Square.

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Other ways of construction (Figure b):

To construct a square within a square where the vertices are not midpoints:

1. Mark a point at a distance '$a$' from each vertex on every side in the same direction (clockwise or anti-clockwise).

2. Let the side of the outer square be $s$. The points will divide the sides into segments '$a$' and '$s-a$'.

3. Joining these four points will always result in an inner Square because the four resulting corner triangles will be congruent by SAS ($a$, $90^\circ$, $s-a$).

Geometric Reasoning:

1. A quadrilateral with four equal sides is by definition a Rhombus.

2. One of the properties of a rhombus is that its opposite angles are equal and its adjacent angles are supplementary (sum to $180^\circ$).

3. Let the quadrilateral be $ABCD$ where $AB = BC = CD = DA$ and $\angle A = 90^\circ$.

4. Then, $\angle C = \angle A = 90^\circ$ (Opposite angles are equal).

5. Also, $\angle A + \angle B = 180^\circ$ (Adjacent angles are supplementary).

$90^\circ + \angle B = 180^\circ \implies \angle B = 90^\circ$.

6. Similarly, $\angle D = 180^\circ - \angle A = 90^\circ$.

7. Since all four sides are equal and all four angles are $90^\circ$, the quadrilateral satisfies the definition of a square.

Conclusion: Yes, a quadrilateral with four equal sides and one $90^\circ$ angle is always a Square.

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Construction and Measurement:

1. Draw a line segment $AB = 4 \text{ cm}$.

2. At $A$, construct an angle of $90^\circ$ and cut an arc of $4 \text{ cm}$ to find point $D$.

3. From $B$ and $D$, draw arcs of $4 \text{ cm}$ each to find the intersection point $C$.

4. Upon joining $BC$ and $CD$ and measuring, we find that all angles are exactly $90^\circ$ and all sides are $4 \text{ cm}$. This confirms the figure is a square.

To Find: The type of quadrilateral where opposite sides are equal.

Given: A quadrilateral $ABCD$ where $AB = CD$ and $BC = DA$.

Construction: Draw a diagonal $AC$.

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Proof:

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

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

By CPCT (Corresponding Parts of Congruent Triangles):

From equation (i), we see that the alternate interior angles are equal for line segments $AB$ and $CD$ with transversal $AC$. Thus, $AB \parallel CD$.

Similarly, from equation (ii), the alternate interior angles for $BC$ and $DA$ are equal. Thus, $BC \parallel DA$.

Conclusion: Since both pairs of opposite sides are parallel and equal, the quadrilateral is a Parallelogram.

Geometric Reasoning:

The figure shown is a concave quadrilateral $ABCD$ (where one internal angle, at $D$, is reflex). To find the angle sum, we can divide the quadrilateral into two triangles by drawing a diagonal $BD$ (internally) or extending a side.

1. Draw diagonal $BD$. This splits the quadrilateral into $\triangle ABD$ and $\triangle BCD$.

2. We know the sum of angles in any triangle is $180^\circ$.

3. Adding equations (i) and (ii):

$(\angle A + \angle ABD + \angle CBD + \angle C) + (\angle ADB + \angle CDB) = 180^\circ + 180^\circ$

$\angle A + \angle B + \angle C + \text{reflex } \angle D = 360^\circ$

Conclusion: Yes, the sum of the internal angles of a concave quadrilateral is also $360^\circ$.

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By Construction and Measurement:

1. Construct a similar figure where $\angle A = 40^\circ, \angle B = 30^\circ, \angle C = 40^\circ$.

2. Measure the reflex angle at $D$ using a protractor. It will measure $250^\circ$.

3. Calculate the sum: $40^\circ + 30^\circ + 40^\circ + 250^\circ = 360^\circ$.

This confirms that the angle sum property holds true for all quadrilaterals.

(i) A quadrilateral whose diagonals are equal and bisect each other must be a square.

False. A quadrilateral whose diagonals are equal and bisect each other is a Rectangle. For it to be a square, the diagonals must also be perpendicular to each other.

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(ii) A quadrilateral having three right angles must be a rectangle.

True. The sum of angles in a quadrilateral is $360^\circ$. If three angles are $90^\circ$, the fourth angle must be $360^\circ - (90^\circ + 90^\circ + 90^\circ) = 90^\circ$. A quadrilateral with all four angles as $90^\circ$ is a Rectangle.

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(iii) A quadrilateral whose diagonals bisect each other must be a parallelogram.

True. By definition, if the diagonals of a quadrilateral bisect each other, it is a Parallelogram.

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(iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus.

False. A Kite also has perpendicular diagonals, but it is not necessarily a rhombus (since all four sides of a kite are not necessarily equal).

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(v) A quadrilateral in which the opposite angles are equal must be a parallelogram.

True. This is a fundamental property. If opposite angles are equal, the adjacent angles become supplementary, implying the opposite sides are parallel.

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(vi) A quadrilateral in which all the angles are equal is a rectangle.

True. If all angles are equal, each must be $360^\circ / 4 = 90^\circ$. An equiangular quadrilateral is a Rectangle.

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(vii) Isosceles trapeziums are parallelograms.

False. A parallelogram must have two pairs of parallel sides. An isosceles trapezium has only one pair of parallel sides.