You can Download KSEEB Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3 Questions and Answers helps you to revise the complete syllabus.

## KSEEB Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 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 of a parallelogram are equal)

(ii) ∠DCB = ∠DAB (opposite angles of a || gm are equal)

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

(iv) m∠DAB + m∠CDA = 180° (Sum of cointerior angles of a y gm between AB y CD)

Question 2.

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

Solution:

(i) ∠D = ∠B (opposite angles of a || gm are equal)

∴ y = 100° [∠B = 100°]

In || gm ABCD

∠y + ∠z = 180° (Sum of cointerior angles between AB || CD)

⇒ 100° + ∠z = 180°

⇒ ∠z = 180° – 100°

⇒ ∠z = 80°

∠z = ∠x (opposite angles of || gm are equal)

∴ ∠x = 80°

Hence ∠A = ∠z = 80°, ∠D = ∠y = 100°, ∠C = ∠x = 80°

(ii) NEST is a parallelogram

∴ ∠N = ∠TSE (opposite angles of a || gm)

⇒ 50° = ∠TSE

∠TSE + ∠z = 180° (Linear pair)

⇒ 50° + ∠z = 180°

⇒ ∠z = 180° – 50°

⇒ ∠z = 130°

Now, in ygm NEST

∠N + ∠E + ∠S + ∠T = 360°

⇒ 50° + y + 50° + x = 360°

⇒ x + y + 100° = 360°

⇒ x + y = 360° – 100°

⇒ x + y = 260°

But ∠x = ∠y (opposite angles of a || gm)

⇒ x + x = 260°

⇒ 2x = 260°

⇒ x = 130°

∠N = 50°, ∠E = y = 130°, ∠S = 50°, ∠T = x = 130°, z = 130°

(iii) ∠AOD = ∠BOC (Vertically opposite angles)

∠AOD = 90°

In ∆AOD

∠OAD + ∠AOD + ∠ADO = 180°

⇒ y + x + 30° = 180°

⇒ y + 90° + 30° = 180°

⇒ y = 180° – 120° = 60°

∠OAD = ∠OCB [AD y BC, hence, alternate angles]

⇒ 60° = ∠OCB

⇒ 60° = z

(iv) SALE is a || gm

∴ ∠S = ∠L (opposite angles of a || gm)

y = 80°

∠SED = ∠L (corresponding angles)

z = 80°

∠A + ∠L = 180° (coniterior angles between AS || LE)

⇒ x + 80° = 180°

⇒ x = 180° – 80° = 100°

∴ x = 100°, y = 80°, z = 80°

(v) ABCD is a ygm

∠A = ∠C (opposite angles of a || gm)

⇒ y = 112°

In ∆ABD,

∠A + ∠ABD + ∠ADB = 180°

⇒ 112° + x + 40° = 180°

⇒ 152° + x = 180°

⇒ x = 180° – 152°

∴ x = 28°

∠ABD = ∠BDC (alternate angles)

⇒ x = z

⇒ 28° = z

∴ x = 28°, y = 112°, 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) It can be a parallelogram in the case of rectangle and square. Since the sum of opposite angles in both of these is 180° and these are parallelograms as well. But in another case when a quadrilateral is a parallelogram, it need not be, that sum of opposite angles is 180°.

(ii) No, it is not a parallelogram. In a parallelogram both pairs of opposite sides are equal. But here AB ≠ DC and AD ≠ BC.

(iii) ∠A ≠ ∠C

∴ It is not a parallelogram.

In parallelogram opposite angles are equal.

Question 4.

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

Solution:

The figure of the kite, in which ∠B = ∠D, but it is not a parallelogram.

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 the ratio be x.

∴ angle measure 3x and 2x

Since, these angles are adjacent

∴ Let ∠D = 3x and ∠A = 2x

But ∠A + ∠D = 180° (Sum of adjacent angles in a || gm)

⇒ 2x + 3x = 180°

⇒ 5x = 180°

∴ x = \(\frac{180^{\circ}}{5}\) = 36°

∴ ∠A = 2 x 36° = 72°

∠D = 3 x 36° = 108°

But ∠D = ∠B and ∠A = ∠C

∴ 108° = ∠B and 72° = ∠C

Question 6.

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

Solution:

ABCD is a parallelogram

∠A = ∠D = x (say) [∵ Adjacent angles are equal]

But ∠A + ∠D = 180° (Sum of adjacent angles in a || gm is 180°)

⇒ x + x = 180°

⇒ 2x = 180°

∴ x = 90°

∴ ∠A = 90° = ∠D

But ∠A = ∠C and ∠B = ∠D

∴ ∠A = ∠B = ∠C = ∠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:

HOPE is a parallelogram

70° + ∠HOP = 180° (linear pair)

⇒ ∠HOP = 180° – 70°

⇒ ∠HOP = 110°

∠E = ∠HOP (opposite angles of a || gm)

⇒ ∠E = 110°

∴ x = 110°

In ∆EPH,

∠E + ∠EHP + ∠EPH = 180°

⇒ x + 40° + ∠EPH = 180°

⇒ 110° + 40° + ∠EPH = 180°

⇒ ∠EPH = 180° – 150°

∴ ∠EPH = 30°

∠EPH = ∠PHO (Alternate angls)

⇒ 30° = ∠PHO

∴ z = 30°

In ∆OPH,

z + y + ∠HOP = 180°

⇒ 30° + y + 110° = 180°

⇒ y = 180° – 140°

∴ y = 40°

∴ x = 110°, y = 40°, z = 30°

Question 8.

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

Solution:

GUNS is a parallelogram

GU = SN (opposite sides of a || gm are equal)

⇒ 3y – 1 = 26

⇒ 3y = 26 + 1

⇒ 3y = 27

∴ y = 9

Now, GS = NU

⇒ 3x = 18

⇒ x = 6

(ii) RUNS is a parallelogram.

SU and RN are diagonals.

Since, diagonals of a parallelogram bisect each other.

∴ OS = OU

⇒ 20 = y + 7

⇒ 20 – 7 = y

⇒ y = 13

Again OR = ON

⇒ 16 = x + y

⇒ 16 = x + 13

⇒ 16 – 13 = x

∴ 3 = x

Question 9.

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

Solution:

RISK is a parallelogram

∠K = ∠RIS (opposite angles of a || gm)

120° = ∠RIS

∠KSI + ∠RIS = 180° (Sum of adjacent angles in a || gm)

⇒ ∠KSI + 120° = 180°

⇒ ∠KSI = 180° – 120°

∴ ∠KSI = 60°

Again, CLUE is a parallelogram

∠CEU = ∠L (opposite angles of a || gm)

⇒ ∠CEU = 70°

Now, in ∆OES

∠O + ∠E + ∠S = 180°

⇒ x + 70° + 60° = 180°

⇒ x + 130° = 180°

⇒ x = 180° – 130°

∴ x = 50°

Question 10.

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

Solution:

In quadrilateral

∠L + ∠M = 180° (Given)

But, these are adjacent angles and their sum is 180°.

Hence, MN || KL

So, KLMN is a trapezium.

In a trapezium, one pair of opposite sides of a quadrilateral are parallel.

Question 11.

Find m∠C in Fig if \(\overline{A B}\) y \(\overline{D C}\)

Solution:

ABCD is a quadrilateral and AB || CD

∴ m∠C + m∠B = 180°(Sum of angles between two parallel lines)

⇒ m∠C + 120° = 180°

⇒ m∠C = 180° – 120°

∴ m∠C = 60°

Question 12.

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

Solution:

PQRS is a quadrilateral in which ∠Q = 130° and ∠R = 90° and SP || RQ

∠P + ∠Q = 180° (Sum of adjacent angles between two parallel lines i.e., SP || RQ)

⇒ ∠P + 130° = 180°

⇒ ∠P = 180° – 130°

∴ ∠P = 50°

Similarly, ∠S + ∠R = 180°

⇒ ∠S + 90° = 180°

⇒ ∠S = 180° – 90°

∴ ∠S = 90°