**Download NCERT Solutions for Class 8 Maths Chapter 3 Exercise 3.3 – Understanding Quadrilaterals**

**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 of a parallelogram are equal)

(ii) ∠DCB = ∠DAB (Opposite angles of a parallelogram are equal)

(iii) OC = OA (Diagonals of a parallelogram bisect each other)

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

**2. Consider the following parallelograms. Find the values of the unknowns x, y, z.**

**i. **

**Solution:**

y = 100° (Opposite angles of a parallelogram are equal)

z + 100° = 180° (Adjacent angles of a parallelogram are supplementary)

z = 80°

x = z = 80° (Opposite angles of a parallelogram are equal)

Therefore, x = 80°, y = 100° and z = 80°

**ii.**

**Solution:**

∴Adjacent angles of a parallelogram are supplementary

y + 50° = 180°

y = 130°

x = y = 130° (Opposite angles of a parallelogram are equal)

z = x = 130° (Corresponding angles)

Therefore, x = y = z = 130°

**iii. **

**Solution:**

x = 90° (Vertically opposite angles)

By Using Angle sum property of a triangle

y + x + 30° = 180°

y + 90° + 30° = 180°

y + 120° = 180°

y = 60°

z = y = 60° (Alternate interior angle)

Therefore, x = 90°, y = 60° and z = 60°

**iv. **

**Solution:**

y = 80° (Opposite angles of a parallelogram are equal)

∴Adjacent angles of a parallelogram are supplementary

x + 80° = 180°

x = 100°

z = 80° (Corresponding angles)

Therefore, x = 100°, y = 80° and z = 80°

**v. **

**Solution:**

y = 112° (Opposite angles of a parallelogram are equal)

∴Adjacent angles of a parallelogram are supplementary

(z + 40°) + 112° = 180°

z + 152° = 180°

z = 28°

x = z = 28° (Alternate interior angles)

Therefore, x = 28°, y = 112° and z = 28°

**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) Quadrilateral ABCD may or may not be a parallelogram with this property.

(ii) No, in a parallelogram, opposite sides must be equal. Here, AD ≠ BC

(iii) No, in a parallelogram, opposite angles are equal. Here ∠A ≠∠C

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

**Solution:**

A kite, for example, has exactly two opposite angles of equal measure.

**5. The measure 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 the two angles be 3x and 2x

∵ Adjacent angles of a parallelogram are supplementary

∴ 3x + 2x = 180°

5x = 180°

x = 36°

One Angle = 3x = 3 x 36° = 108°

Another Angle = 2x = 2 x 36° = 72°

Hence, the measure of angles are 108° and 72°

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

**Solution:**

Let each of two adjacent angle be x

As we know that, Adjacent angles of a parallelogram are supplementary

x + x = 180°

2x = 180°

x = 90°

The remaining two angles will also be 90° because opposite angles of a parallelogram are equal.

Therefore, all angles of the parallelogram are equal to 90°.

**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:**

∠EHO = 70° (Corresponding angles)

40° + z = 70°

z = 30°

∴Adjacent angles of a parallelogram are supplementary

∠EHO + ∠HEP = 180°

(40° + 30°) + x = 180°

70° + x = 180°

x = 110°

∠HPO = ∠EHP (Alternate interior angles)

y = 40°

Therefore, x = 110°, y = 40° and z = 30°

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

**i.**

**Solution:**

SG = NU and GU = SN (Opposite sides of a parallelogram are equal)

3x = 18

x = 6

3y – 1 = 26

3y = 27

y = 9

Therefore, x = 6 and y = 9

**ii. **

**Solution:**

y + 7 = 20 and x + y = 16 (Diagonals of a parallelogram bisect each other)

y + 7 = 20

y = 13

x + y = 16

x + 13 = 16

x = 3

Therefore, x = 3 and y = 13

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

**Solution:**

Let the point of intersection of EC and IS be P

In parallelogram RISK

∴Adjacent angles of a parallelogram are supplementary

∠K + ∠S = 180°

120° + ∠S = 180°

∠S = 60°

In parallelogram CLUE

∠E = ∠L (Opposite angles of a parallelogram are equal)

∠E = 70°

By using Angle sum property of a triangle, In △SEP

∠E + ∠S + ∠O = 180°

70° + 60° + x = 180°

130° + x = 180°

x = 50°

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

**Solution:**

∠NML + ∠MLK = 100° + 80° = 180°

Since, adjacent angles are supplementary

Therefore, MN ∥ LK

Hence, NMLK is a trapezium.

**11. Find m∠C in the given figure if || .**

**Solution:**

∵ ||

∴ ∠B + ∠C = 180° (Adjacent angles)

120° + ∠C = 180°

m∠C = 60°

**12. Find the measure of ∠P and ∠S if || in the given figure.
(If you find m∠R, is there more than one method to find m∠P?)**

**Solution:**

∵ ||

∴ ∠S+ ∠R = 180° (Adjacent angles)

∠S + 90° = 180°

∠S = 90°

∠P+ ∠Q = 180° (Adjacent angles)

∠P + 130° = 180°

∠P = 50°

∠P can also be found by using the angle sum property of a quadrilateral.

**Download NCERT Solutions for Class 8 Maths Chapter 3 Exercise 3.3 – Understanding Quadrilaterals**

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