**NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals
Ex 3.3**

NCERT Solutions for Class
8 Maths Chapter 3
Understanding Quadrilaterals Ex 3.3 are the
part of NCERT Solutions for Class 8 Maths. Here you can find the NCERT
Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3**.**

**Ex 3.3 Class 8 Maths 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 = ……..

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

(ii) ∠DCB = ∠BAD [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]

**Ex 3.3 Class 8 Maths Question 2.**

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

(i) ABCD is a parallelogram.

∠B = ∠D [Opposite angles of a parallelogram are equal]

∠D = 100°

⇒ y = 100°

∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary]

⇒ z + 100° = 180°

⇒ z = 180° – 100° = 80°

∠A = ∠C [Opposite angles of a parallelogram are equal]

x = 80°

Hence, x = 80°, y = 100° and z = 80°

(ii) PQRS is a parallelogram.

∠P + ∠S = 180° [Adjacent angles of parallelogram are supplementary]

⇒ x + 50° = 180°

⇒ x = 180° – 50° = 130°

Now, ∠P = ∠R [Opposite angles of a parallelogram are equal]

⇒ x = y

⇒ y = 130°

Also, y = z [Alternate angles]

z = 130°

Hence, x = 130°, y = 130° and z = 130°

(iii) ABCD is a rhombus.

x = 90° [∵ Diagonals bisect each other at 90°]

Now, in ∆OCB,

x + y + 30° = 180° (Angle sum property)

⇒ 90° + y + 30° = 180°

⇒ y + 120° = 180°

⇒ y = 180° – 120° = 60°

y = z (Alternate angles)

⇒ z = 60°

Hence, x = 90°, y = 60° and z = 60°.

(iv) ABCD is a parallelogram.

∠A + ∠B = 180° (Adjacent angles of a parallelogram are supplementary)

⇒ x + 80° = 180°

⇒ x = 180° – 80° = 100°

Now, ∠D = ∠B [Opposite angles of a parallelogram are equal]

⇒ y = 80°

Also, z = ∠B = 80° (Corresponding angles)

Hence, x = 100°, y = 80° and z = 80°

(v) ABCD is a parallelogram.

∠D = ∠B [Opposite angles of a parallelogram are equal]

y = 112°

x + y + 40° = 180° [Angle sum property]

⇒ x + 112° + 40° = 180°

⇒ x + 152° = 180°

⇒ x = 180° – 152 = 28°

z = x = 28° (Alternate angles)

Hence x = 28°, y = 112°, z = 28°.

**Ex 3.3 Class 8 Maths 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) For
∠D + ∠B = 180°,
quadrilateral ABCD may be a parallelogram if the following conditions are also
fulfilled.

(a) The sum of measures of adjacent angles should be 180°.

(b) Opposite angles should also be of same measures.

So,
ABCD can be a parallelogram, but need not be a parallelogram.

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

In a parallelogram, opposite sides are equal.

Here, AD ≠ BC

Thus, ABCD cannot be a parallelogram.

(iii) ∠A = 70° and ∠C = 65°

In a parallelogram, opposite angles are equal.

Here, ∠A ≠ ∠C

Hence, ABCD is not a parallelogram.

**Ex 3.3 Class 8 Maths Question 4.**

Draw a rough figure of a quadrilateral that is not a
parallelogram but has exactly two opposite angles of equal measure.**Solution:
**Here, ABCD is a rough figure of a quadrilateral in
which m ∠A = m ∠C but it is not a parallelogram. It is a kite.

**Ex 3.3 Class 8 Maths 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 is a parallelogram such that ∠B : ∠C = 3 : 2.

∠B + ∠C = 180° (Adjacent angles of a parallelogram are supplementary)

3x + 2x = 180°

⇒ 5x = 180°

⇒ x = 36°

Thus, ∠B = 3 × 36 = 108°

∠C = 2 × 36° = 72°

∠B = ∠D = 108°

and ∠A = ∠C = 72°

Hence, the measures of the angles of the parallelogram are

108°, 72°, 108° and 72°.

**Ex 3.3 Class 8 Maths 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 in which ∠A = ∠B

∠A + ∠A = 180°

⇒ 2∠A = 180°

⇒ ∠A = 90°

Thus, ∠A = ∠C = 90° and ∠B = ∠D = 90°

[Opposite angles of a parallelogram are equal]

**Ex 3.3 Class 8 Maths 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 = 40° (Alternate
angles)

∠z + 40° = 70° (Exterior angle property)

⇒ ∠z = 70° – 40° = 30°

z = ∠EPH (Alternate angle)

In ∆EPH,

∠x + 40° + ∠z = 180° (Angle sum property)

⇒ ∠x + 40° + 30° = 180°

⇒ ∠x + 70° = 180°

⇒ ∠x = 180° – 70° = 110°

Hence x = 110°, y = 40° and z = 30°.

**Ex 3.3 Class 8 Maths Question 8.**

The
following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are
in cm)**Solution:****
**(i) GU
= SN (Opposite sides of a
parallelogram)

⇒ y = 20 – 7 = 13

Also, ON = OR

⇒ x + y = 16

⇒ x + 13 = 16

⇒ x = 16 – 13 = 3

Hence, x = 3 cm and y = 13 cm.

**Ex 3.3 Class 8 Maths Question 9.**

In the above figure, both RISK and CLUE are parallelograms. Find the value of
x.**Solution:
**Here, RISK and CLUE are two parallelograms.

∠K + ∠2 = 180° (Adjacent angles of a parallelogram are supplementary)

120° + ∠2 = 180°

∠2 = 180° – 120° = 60°

In ∆OES,

∠x + ∠1 + ∠2 = 180° (Angle sum property)

⇒ ∠x + 70° + 60° = 180°

⇒ ∠x + 130° = 180°

⇒ ∠x = 180° – 130° = 50°

Hence, x = 50°

**Ex 3.3 Class 8 Maths Question 10.**

Explain how this figure is a trapezium. Which of its
two sides are parallel?**Solution:
**∠M + ∠L = 100° + 80° = 180°

∠M and ∠L are the adjacent angles, and the sum of the adjacent interior angles is 180°. Therefore, KL is parallel to NM

Hence, KLMN is a trapezium.

**Ex 3.3 Class 8
Maths Question 11.**

Find m ∠C in the following figure if AB || DC.**Solution:
**Given: AB || DC

m ∠B + m ∠C = 180° (Co-interior angles are supplementary)

120° + m ∠C = 180°

m ∠C = 180° – 120° = 60°

Hence, m ∠C = 60°

**Ex 3.3 Class 8
Maths Question 12.**

Find the measure of ∠P and ∠S if SP || RQ in figure. (Is there more than one method to
find m ∠P?)**Solution:
**Given: m ∠Q = 130° and m ∠R = 90°

SP || RQ (Given)

∠P + ∠Q = 180° (Co-interior angles are supplementary)

⇒ ∠P + 130° = 180°

⇒ ∠P = 180° – 130° = 50°

and, ∠S + ∠R = 180° (Co-interior angles are supplementary)

⇒ ∠S + 90° = 180°

⇒ ∠S = 180° – 90° = 90°

**Alternate Method:**

∠Q = 130°, ∠R = 90° and ∠S = 90°

We know that

∠P + ∠Q + ∠R + ∠Q = 360° (Angle sum property of quadrilateral)

⇒ ∠P + 130° + 90° + 90° = 360°

⇒ ∠P + 310° = 360°

⇒ ∠P = 360° – 310° = 50°

Hence, m ∠P = 50°

Thus, there are more
than one method to find m ∠P.

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