These NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3

Question 1.

Given a parallelogram ABCD. Complete each statement along with the definition or property used.

(i) AD = __________

(ii) ∠DCB = __________

(iii) OC = __________

(iv) m∠DAB + m∠CDA = __________

Solution:

(i) AD = BC (opposite sides are equal)

(ii) ∠DCB = ∠DAB (opposite angles are equal)

(iii) OC = OA (Diagonals bisect each other)

(iv) m∠DAB + m∠CDA = 180° (Adjacent angles are supplementary)

Question 2.

Consider the following parallelograms. Find the values of the unknown x, y, z.

Solution:

(i) ∠y = 100° (opposite angles of a parallelogram are equal)

∠x + 100° = 180° (adjacent angles in a parallelogram)

∠x = 180° – 100° = 80°

∠z = ∠x = 80° (opposite angles of parallelogram are equal)

∴ ∠x = 80°, ∠y = 100°, ∠z = 80°

(ii) Opposite angles are equal.

∠1 = 50° (opposite angles are equal)

∠1 + ∠z = 180° (Linear pair)

50 + ∠z = 180

∠z = 180° – 50°= 130°

x + ∠1 + y + 50° = 360°

⇒ x + 50° + y + 50° = 360°

⇒ x + y + 100° = 360°

⇒ x + y = 360° – 100°

⇒ x + y = 260°

∴ x = y (opposite angles of a parallelogram)

x = \(\frac{260^{\circ}}{2}\) = 130°

Thus x = 130°, y = 130° and z = 130°.

(iii) x = 90° (vertically opposite angles are equal)

x + y + 30° = 180°

(sum of the angles of a triangle = 180°)

90° + y + 30° = 180°

y + 120° = 180°

∠y = 180° – 120° = 60°

In the parallelogram ABCD

AD || BC and BD is a transversal.

∴ y = z (alternate angles are equal)

z = y = 60°

Thus x = 90°, y = 60° and z = 60°

(iv) In the parallelogram ABCD

y = 80° (opposite angles are equal)

AD || BC and CD is a transversal equal)

∴ z = 80° (corresponding angles are equal)

x + 80° = 180° (sum of the adjacent angle is 180°)

x = 180° – 80° = 100°

Thus, x = 100°, y = 80° and z = 80°

(v) y = 112° (in a parallelogram, opposite angles are equal)

In ΔACD,

x + y + 40° = 180° (sum of the angles of a triangle is 180°)

⇒ x + 112° + 40° = 180°

⇒ x + 152°= 180°

⇒ x = 180° – 152° = 28°

∠C + ∠B = 180° (sum of the adjacent angles of a parallelogram is 180°)

⇒ 40° + z + 112° = 180°

⇒ z + 152° = 180°

⇒ z = 180° – 152° = 28°

Thus, x = 28°, y = 112° and z = 28°.

Question 3.

Can a quadrilateral ABCD be a parallelogram if

(i) ∠D + ∠B = 180°?

(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?

(iii) ∠A = 70° and ∠C = 65°?

Solution:

(i) In a quadrilateral ABCD

∠D + ∠B = 180° can be, but need not be

∴ The quadrilateral may be a parallelogram but not always

(ii) In a quadrilateral ABCD

AB = DC = 8 cm

AD = 4 cm

BC = 4.4 cm

∴ Opposite sides AD and BC are not equal.

∴ It cannot be a parallelogram.

(iii) In a quadrilateral ABCD

∠A = 70° and ∠C = 65°

∵ Opposite angles ∠A ≠ ∠C

∴ It cannot be a parallelogram.

Question 4.

Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.

Solution:

In the adjoining figure, ABCD is not a parallelogram such that opposite angles ∠B and ∠D are equal. It is a kite.

Question 5.

The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.

Solution:

Let ABCD be a parallelogram in which adjacent angles ∠A and ∠B are 3x and 2x respectively.

Since adjacent angles are supplementary.

∴ ∠A + ∠B = 180°

⇒ 3x + 2x = 180°

⇒ 5x = 180°

⇒ x = \(\frac{180^{\circ}}{5}\) = 36°

∠A = 3 × 36° = 108°

and ∠B = 2 × 36° = 72°

Since opposite angles are equal.

∴ ∠D = ∠B = 72° and ∠C = ∠A = 108°

∴ ∠A = 108°, ∠B = 72°, ∠C = 108° and ∠D = 72°

Question 6.

Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.

Solution:

Let ABCD be a parallelogram such that adjacent angles ∠A = ∠B

Since ∠A + ∠B = 180°

∠A = ∠B = \(\frac{180^{\circ}}{2}\) = 90°

Since opposite angles of a parallelogram are equal

∴ ∠A = ∠C = 90° and ∠B = ∠D = 90°

Thus, ∠A = 90°, ∠B = 90°, ∠C = 90° and ∠D = 90°

Question 7.

The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.

Solution:

y + z = 70°

(An exterior angles of a triangle is equal to the sum of the opposite interior angles)

∵ In ∆HOP

∠HOP = 180° – 70° = 110° (sum of the adjacent angle is 180°)

Now, x = ∠HOP = 110° (opposite angles of a parallelogram are equal)

EH || OP and PH is a transversal

∴ y = 40° (alternate angles are equal)

From (1),

y + z = 70°

⇒ 40° + z = 70°

⇒ z = 70° – 40° = 30°

Thus x = 110°, y = 40° and z = 30°

Question 8.

The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)

Solution:

(i) GUNS is a parallelogram.

GS = NU (opposite sides are equal)

3x = 18

⇒ x = \(\frac{18}{3}\) = 6

In the parallelogram GUNS

GU = SN (opposite sides are equal)

⇒ 3y – 1 = 26

⇒ 3y = 26 + 1

⇒ 3y = 27

⇒ y = \(\frac{27}{3}\) = 9

Thus, x = 6 cm and y = 9 cm.

(ii) RUNS is a parallelogram and the di-agonal RN and US bisect each other.

∴ y + 7 = 20

⇒ y = 20 – 7 = 13

x + y = 16

⇒ x + 13 = 16

⇒ x = 16 – 13 = 3

Thus, x = 3 cm and y = 13 cm

Question 9.

In the figure given below both RISK and CLUE are parallelograms. Find the value of x.

Solution:

RISK is a parallelogram

∠R + ∠K = 180° (adjacent angles of a parallelogram are supplementary)

∴ ∠R + 120° = 180°

∠R = 180° – 120° = 60°

In the parallelogram RISK

∠R = ∠S (opposite angles are equal)

∠S = 60°

CLUE is also a parallelogram.

∠E = ∠L = 70° (opposite angle of a parallelogram)

∠E = 70°

Now, in triangle ESO

∠E + ∠S + x = 180°

⇒ 70° + 60° + x = 180°

⇒ 130° + x = 180°

⇒ x = 180° – 130°

⇒ x = 50°

Question 10.

Explain how this figure is a trapezium. Which of its two sides are parallel?

Solution:

∠KLM + ∠NML = 80° + 100° = 180°

∴ KL || NM (The sum of consecutive interior angles is 180°)

∴ The given figure KLMN is a trapezium.

Question 11.

Find m∠C in the figure given below if AB || DC.

Solution:

AB || DC and BC is a transversal,

∵ m∠B + m∠C = 180°

sum of interior angles is 180°

m∠C = 180° – m∠B

∴ m∠C = 180° – 120° = 60°

Question 12.

Find the measure of ∠P and ∠S if SP || QR in figure (if you find m∠R, is there more than one method to find m∠P)

Solution:

PQRS is a trapezium such that SP || RQ and PQ is a transversal.

∴ m∠P + m∠Q = 180° (Interior angles are supplementary)

m∠P + 130° = 180°

m∠P = 180° – 130° = 50°

Also, m∠S + m∠R = 180°

⇒ m∠S + 90° = 180°

⇒ m∠S = 180° – 90°

⇒ m∠S = 90°

m∠P + m∠Q + m∠R + m∠S = 360° (sum of the angles of a quadrilateral is 360°)

⇒ m∠P + 130° + 90° + 90° = 360°

⇒ m∠P + 130° + 90° + 90° = 360°

⇒ m∠P + 130° + 90° + 90° = 360°

⇒ m∠P + 310° = 360°

⇒ m∠P = 360° – 310°

⇒ m∠P = 50°

Hence, m∠P = 50° and m∠S = 90°