## MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

**MP Board Class 8 Maths Chapter 3 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:

Given that ABCD is a parallelogram.

(i) AD = BC [∵ In a parallelogram opposite sides are equal]

(ii) ∠DCB = ∠DAB [∵ In a parallelogram opposite angles are equal]

(iii) OC = OA [∵ In a parallelogram diagonals bisect each other]

(iv) m∠DAB + m∠CDA = 180° [∵ Adjacent angles in a parallelogram are supplementary]

**Class 8 Maths Chapter 3 MP Board Question 2.**

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

Solution:

(i) ABCD is a parallelogram in which ∠B = 100° (given)

∴ ∠A + ∠B = 180° [∵ Sum of adjacent angles is 180°]

⇒ z +100° = 180°

⇒ z = 180° – 100° = 80°

Also ∠B = ∠D and ∠A = ∠C

[∵ Opposite angles are equal]

∴ ∠B = 100° = ∠D = y and ∠A = z = 80° = ∠C = x

∠ x = 80°, y = 100°, z = 80°.

(ii) y + 50° = 180° [∵ Sum of adjacent angles is 180°]

⇒ y = 180° – 50° = 130°

Also, y = x = 130° [∵ Opposite angles are equal in a parallelogram]

And 180°- z = 50° [Linear pair]

⇒ z = 180°- 50° ⇒ z = 130°

(iii) Clearly, x = 90°

[∵ Vertical opposite angles are equal] Also, x + y + 30° = 180°

[By using angle sum property of a triangle]

⇒ 90° + 30° + y = 180°

⇒ 120° + y = 180° ⇒ y = 180° – 120° = 60°

Since, alternate angles are equal in a parallelogram

∴ y = z = 60°

Thus x = 90°, y = 60° and z = 60°.

(iv) Since x + 80° = 180°

[Sum of adjacent angles is 180°] ⇒ x = 180° – 80° = 100°

Also 80° = y [∵ Opposite angles are equal in a parallelogram]

And x = 180° – z ⇒ 100° = 180° – 2° ⇒ 2 = 180° – 100° = 80°

Thus x = 100°, y = 80° and 2 = 80°.

(v) y = 112° [∵ Opposite angles are equal in a parallelogram]

y + x + 40° = 180°

[By angle sum property of a triangle]

⇒ 112°+ x+ 40° = 180°

⇒ x +152° = 180° ⇒ x = 180° – 152°

⇒ x = 28°

⇒ x = 2 = 28° [∵ Alternate angles are equal in a parallelogram]

**MP Board Class 8th Maths Chapter 3 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) Yes, but need not be true.

(ii) No, because in a parallelogram opposite sides are equal but here AD ≠ BC.

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

**Class 8 MP Board Maths Chapter 3 Question 4.**

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

Solution:

We can draw a figure of a kite in which exactly two opposite angles are equal.

Hence ∠D = ∠B.

**MP Board Class 8 Maths Solutions English Medium Chapter 3 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 given parallelogram.

Let ∠A = 3x and ∠B = 2x.

Since the sum of adjacent angles in a parallelogram is 180°

∴ m∠A + m∠B = 180°

⇒ 3x + 2x = 180°

⇒ 5x = 180° ⇒ x = 36°

∴ m∠A = 3 × 36° = 108° and m∠B = 2 × 36° = 72°

Also, m∠A = m∠C = 108°

[∵ In a parallelogram opposite angles are equal]

and m∠B = m∠D = 72°.

**MP Board Solutions Class 8 Maths Chapter 3 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, let m∠A = m∠B = x

Since sum of two adjacent angles is 180°

∴ m∠A + m∠B = 180°

⇒ x + x = 180° ⇒ 2x = 180°

⇒ x = 90°

Also m∠A = m∠C = 90°

[∵ In a parallelogram opposite angles are equal]

and m∠B = m∠D = 90°.

**MP Board Class 8 Maths Chapter 3 Question 7.**

The given figure HOPE is a parallelogram. Find the angle measures x, y and z. State the

properties you use to find them.

Solution:

y = 40°

[Since PO || HE ∴ alternate angles are equal]

70° = y + 2 [Exterior angle property of a triangle]

⇒ z = 70° – 40° = 30°

∠POH = 180° – 70° = 110° [Linear pair]

∠POH = x = 110° [Opposite angles are equal]

Thus x = 110°, y = 40°, 2 = 30°.

**Class 8 Maths Chapter 3 Exercise 3.3 Solutions Question 8.**

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

Solution:

(i) Since GUNS is a parallelogram.

∴ GS = NU and GU = SN

[∵ In a parallelogram opposite sides are equal]

⇒ 3x = 18 ⇒ x = 6

and 3y -1 = 26 ⇒ 3y = 1 + 26 = 27

⇒ y = 9

Thus x = 6 cm and y = 9 cm.

(ii) Since diagonals bisects each other in a parallelogram.

So, 20 = y + 7 ⇒ y = 13

Also, x + y= 16 ⇒ x = 16 – 13 = 3

Thus x = 3 cm and y = 13 cm.

**MP Board Class 8 Maths Chapter 3 Exercise 3.1 Question 9.**

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

Solution:

Since RISK and CLUE are parallelograms.

∴ ∠SKR = ∠RIS = 120° [∵ Opposite angles are equal]

Also, ∠ULC = ∠UEC = 70° [Opposite angles are equal]

∠RIS + ∠ISK = 180° [Adjacent angles are supplementary]

⇒ ∠ISK = 180° – 120° = 60°

In ∆OES, we have

70° + x + 60° = 180° [Angle sum property]

⇒ x + 130° = 180°

⇒ x = 180° – 130° = 50°.

**MP Board Class 8 Math Chapter 3 Question 10.**

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

Solution:

In a trapezium, only one pair of opposite sides are parallel, whereas other pair of opposite sides are non-parallel.

∴ KLMN is a trapezium because MN || KL.

[∵ Sum of two adjacent interior angles is 180° = ( 80° + 100°)].

**Maths Class 8 Chapter 3 Exercise 3.3 Question 11.**

Find m∠C in given figure if AB || DC.

Solution:

We are given, \(\overline{A B} \| \overline{D C}\)

∴ Sum of two adjacent interior angles is 180°

i. e., ∠B + ∠C = 180° ⇒ 120° + ∠C= 180°

⇒ ∠C = 180° -120° = 60°.

Thus ∠C = 60°.

**Class 8 Subject Maths Exercise 3.3 Question 12.**

Find the measure of ∠P and ∠S if \(\overline{S P} \| \overline{R Q}\) in figure. (If you find m∠R, is there more than one method to find m∠P?)

Solution:

Since \(\overline{S P} \| \overline{R Q}\)

Thus ∠R + ∠S = 180°

[∵ Sum of two adjacent interior angles is 180°]

⇒ 90° + ∠S = 180° ⇒ ∠S = 180° – 90° = 90°

∠P + ∠Q + ∠R + ∠S = 360° [∵ Sum of all angles of a quadrilateral is 360°]

⇒ ∠P + 310° = 360°

⇒ ∠P = 360° – 310°

⇒ ∠P = 50°

Thus ∠P = 50° and ∠S = 90°.

Also m∠P can be found as ∠P + ∠Q = 180°

[Adjacent angles are supplementary]

⇒ ∠P + 130° = 180°

⇒∠P = 180° -130° ⇒ ∠P = 50°