Haryana State Board HBSE 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions 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°.

[Interior opposite angles-since \(\overline{\mathrm{AB}}\) || DC]

Question 2.

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

Solution:

(i) ∠D = ∠B

[Opposite angles are equal]

∴ y = 100°

[∵ ∠B = 100°]

In ||gm ABCD,

∠A + ∠B + ∠C + ∠D = 360°

∠z + 100° + ∠x + 100° = 360°

∠z + ∠x + 200 = 360°

∴ ∠z + ∠x = 360 – 200 = 160°

But, ∠z = ∠x

[Opposite angles are equal]

2∠z = 160°

∴ ∠z = 80° = ∠x

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

(ii) ABCD is a parallelogram

∴ ∠D = ∠ABC

[Opposite angles are equal]

∴ ∠ABC = 50°

∠ABC + ∠CBP = 180° [Linear pair]

50° + ∠CBP = 180°

∴ ∠CBP = 180° – 50° = 130°

∴ ∠CBP = ∠ = 130°

In ||gm ABCD,

∠A + ∠B + ∠C + ∠D = 360°

x + 50° + y + 50° = 360°

x + y + 100° = 360°

x + y = 360° – 100° = 260°

But, ∠x = ∠y

[Opposite angles are equal]

2∠x = 260°

∴ ∠x = 130° = ∠y

Hence, ∠x = 130°; ∠z = 130°; ∠y = 130°.

(iii) ∠AOD = ∠BOC

Here, ∠BOC = 90°

∴ ∠AOD = 90°

In △AOD,

∠AOD + ∠ADO + ∠OAD = 180°

[Angle sum property]

90° + 30° + ∠OAD = 180°

120° + ∠OAD = 180°

∴ ∠OAD = 180° – 120° = 60°

∴ ∠OAD = ∠y = 60°

∠y = ∠z

[∵ AD || CD] (alt. int. angle]

∴ ∠z = 60°.

(iv) ABCD is a ||gm

∴ ∠B = ∠D

∴ ∠y = 80° [∵ ∠D = y]

∠B = ∠z

[Corresponding angles since AB || BP]

∴ ∠z = 80°

∠z + ∠BCD = 180° [Linear pair]

80° + ∠BCD = 180°

∴ ∠BCD = 180° – 80° = 100°

∠BCD = ∠x [Opposite angles are equal]

∴ ∠x = 100°

Hence, ∠x = 100°, ∠z = 80°, ∠y = 80°.

(v) ∠B = ∠D

[Opposite angles are equal]

∴ ∠D = 112°

InABCD, ∠A + ∠B + ∠C + ∠D = 360°

or ∠A + 112° + ∠C + 112° = 360°

or ∠A + ∠C + 2240 = 360°

or ∠A + ∠C = 360° – 224°

or ∠A + ∠C = 136°

But ∠A = ∠C

∴ 2∠A = 136°

∠A = 68°

∴ ∠A = ∠C = 68°

∠A = 40°+ z

68° = 40° + z

∴ ∠ = 68° – 40° = 28°

∴ ∠z = ∠x = 28° [Alternative angles]

Hence, 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) Can be, but need not be, because sum of opposite angles of a rectangle and a square is 180°, but it need not be a parallelogram, rhombus.

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

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

Question 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

Let ∠B = ∠D = x [Opposite angles property]

In parallelogram ABCD,

∠A + ∠B + ∠C + ∠D = 360°

75° + x + 75° + x = 360°

or 2x +150° = 360°

or 2x = 360°-150° = 210°

x = 105°

Hence, ∠A = 75°, ∠B = 105°, ∠C = 75° and ∠D = 105°.

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:

Given: ABCD is a parallelogram, ∠A and ∠B are in the ratio of 3 : 2.

Find ∠A, ∠B, ∠C and ∠D.

Let the ratio = x

∴ ∠A = 3x, ∠B = 2x

∠A + ∠B = 180° [Sum of interior angles]

or 3x + 2x = 180°

or 5x = 180°

∴ x = 36°

∠A = 3 × 36° = 108°

∠B = 2 × 36° = 72°

∠A = ∠C = 108°

∠B = ∠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:

Given : ∠A = ∠B = x let

∠A + ∠B = 180°

[Sum of interior angles]

∵ AD | | BC

x + x = 180°

2x = 180°

∴ x = 90°

Question 7.

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

Solution:

70° + ∠POH = 180° [Linear pair]

∴ ∠POH = 180° -70°

∴ ∠POH = 110°

∠POH = ∠HEP

[Opposite angles of a parallelogram HOPE]

∴ x = 110°

∠EHP = ∠HPO

[Alternative interior angles]

∴ ∠y = 40°.

Question 8.

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

Solution:

(i) GUNS is a parallelogram

∴ GU = SN

[Opposite sides of a ||gm]

or 3y – 1 = 26

3y = 26 + 1

3y = 27

∴ y = \(\frac{27}{3}\) = 9

GS = UN

[Opposite sides of a ||gm]

3x = 18

∴ x = 6

Therefore, x = 6, y = 9.

(ii) y + 7 = 20

y = 20 – 7 = 13

∴ y = 13

x + y = 16

[Diagonals of a ||gm bisect each other]

or x + 3 = 16

∴ x = 16 – 13 = 3

Therefore, x = 3, y = 13.

Question 9.

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

Solution:

RISK is a ||gm

∴ ∠K = ∠RIS = 120°

In ||gm RISK,

∠KRI = ∠ISK = x (let)

∠SKR + ∠KRI + ∠RIS + ∠ISK = 360°

or 120° + x + 120° + x = 360°

or 240° + 2x = 360°

or 2x = 360° – 240°

= 120°

x = \(\frac{120}{2}\) = 60°

∴ ∠ISK = 60°

CLUE is a ||gm

∴ ∠L = ∠CEU = 70

In triangle shape,

∠x + ∠CEU + ∠ISK = ∠x + ∠S + ∠E

= 180°

x + 70° + 60°

= 180°[∵ ∠x = 50°]

∴ x = 180° – 70° – 60°

= 50°.

Question 10.

Explain how this figure is a trapezium. Which of its two sides are parallel ? (Fig. 3.29)

Solution:

∠KLM + ∠NML = 80° + 100° = 180° [Sum of interior opposite angles 180°]

∴ \(\overline{\mathrm{NM}}\) || \(\overline{\mathrm{KL}}\). So KLMN is a trapezium.

Question 11.

Find m∠C in Fig. 3.30, if \(\overline{\mathrm{AB}}\). \(\overline{\mathrm{DC}}\).

Solution:

∠B + ∠C = 180°

120°+ ∠C = 180°

∴ ∠C = 180° – 120° = 60° ^

∴ ∠C = 60°.

Question 12.

Find the measure of ∠P and ∠S if \(\overline{\mathrm{SP}}\).\(\overline{\mathrm{RQ}}\) in Fig. 3.31. (If you find m∠R, is there more than one method to find m∠P)?

Solution:

∠QRS + ∠RSP = ,180°

[Sum of interior opposite angle is 180° because \(\overline{\mathrm{SP}}\).\(\overline{\mathrm{RQ}}\)]

90° + ∠RSP = 180°

∴ ∠RSP = 90°

In quad. PQRS,

∠P + ∠Q + ∠R + ∠S = 360°

∠P + 130° + 90° + 90° = 360°

or ∠P + 310° = 360°

∴ ∠P = 360° – 310° = 50°

∴ ∠P = 50°; ∠S = 90°