## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.3

Question 1.

Identify all the quadrilaterals that have

(i) four sides of equal length

(ii) four right angles.

Solution:

(i) Any quadrilateral whose four sides are equal

in length is a square or rhombus.

(ii) A quadrilateral having four right angles

is a square or a rectangle.

Question 2.

Explain how a square is

(i) a quadrilateral

(ii) a parallelogram

(iii) a rhombus

(iv) a rectangle.

Solution:

(i) A square is a quadrilateral which has four sides

and four angles whose sum is 360°.

(ii) A square is a parallelogram whose opposite sides are parallel.

(iii) A square is a parallelogram whose sides are equal

and so, it is a rhombus.

(iv) A square is a parallelogram whose each angle is 90°.

So, it is a rectangle.

Question 3.

Name the quadrilaterals whose diagonals

(i) bisect each other

(ii) are perpendicular bisectors of each other

(iii) are equal.

Solution:

(i) Rectangle, square, rhombus, parallelogram.

(ii) Square, rhombus.

(iii) Square, rectangle.

Question 4.

One of the diagonals of a rhombus and its sides are equal. Find the angles of the rhombus.

Solution:

In a rhombus, side and one diagonal are equal.

∴ Angles will be 60° and 120°.

Question 5.

In the given figure, ABCD is a rhombus, find the values of x, y and z.

Solution:

In rhombus ABCD.

∵ The diagonals of rhombus bisect each other at right angles.

∴ AO = OC and BO = OD

AO = x, OC = 8 cm, BO =y and OD = 6 cm

∴ x = 8 cm and y = 6 cm

In ∆AOB,

AB^{2} = AO^{2} + BO^{2}

AB^{2} = 8^{2} + 6^{2}

AB^{2} = 64 + 36

AB^{2} = 100 = (10)^{2}

AB = 10 cm

Question 6.

In the given figure, ABCD is a trapezium. If ∠A : ∠D = 5 : 7, ∠B = (3x + 11)° and ZC = (5x – 31)°, then find all the angles of the trapezium.

Solution:

In the given figure,

ABCD is a trapezium in which DC || AB

∠A : ∠D = 5 : 7

∠B = (3x + 11)° and

∠C = (5x – 31)°

∵ ABCD is a trapezium

∴ ∠B + ∠C = 180° (Cointerior angle)

3x + 11° + 5x – 31° = 180°

8x – 20° = 180° ⇒ 8x = 180° + 20° = 200°

⇒ x = \(x=\frac{200^{\circ}}{8}=25^{\circ}\)

2x = 180°- 118°

2x = 62° ⇒ x = 31°

∴ ∠ABO = 31°

(ii) Also, ∠AOB + ∠AOD = 180° (Linear pair)

∴ ∠AOD = 180°- 118° = 62°

Now, In ∆AOD, AO = DO = y°

∴ 62 + 2y= 180°

2y = 180° – 62°

2y = 180°- 62°

2y = 118° ⇒ y = 59°

(iii) ∠OCB = ∠OAD = 59° (Alternate angles)

Question 7.

In the given figure, ABCD is a rectangle. If ∠CEB : ∠ECB = 3 : 2 find

(i) ∠CEB,

(ii) ∠DCF

Solution:

In ∆ BCE, ∠B = 90° (∵ ABCD is a rectangle)

∴ ∠CEB + ∠ECB = 90°

3x + 2x = 90°

⇒ x= 18°

∴ ∠CEB = 3x = 3 × 18° = 54°

Now, ∠CEB = ∠ECD = 54° (Alternate angles)

Also ∠ECD + ∠DCF = 180° (Linear pair)

⇒ ∠DCF = 180 – 54= 126°

Question 8.

In the given figure, ABCD is a rectangle and diagonals intersect at O. If ∠AOB = 118°, find

(i) ∠ABO

(ii) ∠ADO

(iii) ∠OCB

Solution:

In ∆ ABO, OA = OB

(∵ diagonals of a rectangle bisect each other)

∴ ∠OAB = ∠OBA = x °

⇒ 118° + x + x = 180°

2x= 180°- 118°

2x = 62° ⇒ x = 31°

∴ ∠ABO = 31°

(ii) Also, ∠AOB + ∠AOD = 180° (Linear pair)

∴ ∠AOD = 180° – 118° = 62°

Now, In ∆ AOD, AO = DO = y°

∴ 62 + 2y = 180°

2y = 180°- 62°

2y = 118° ⇒ y = 59°

(iii) ∠OCB = ∠OAD = 59° (Alternate angles)

Question 9.

In the given figure, ABCD is a rhombus and ∠ABD = 50°. Find :

(i) ∠CAB

(ii) ∠BCD

(iii) ∠ADC

Solution:

(i) We know that diagonals of a rhombus

are ⊥ to each other.

∴ ∠BOA = 90°

In ∆ AOB,

∠OAB + ∠BOA + ∠ABO = 180°

∠OAB + 90° + 50° = 180°

∠OAB = 180 – 140 = 40°

∴ ∠CAB = ∠OAB = 40°

(ii) ∠BCD = 2 ∠ACD = 2 × 40° = 80°

(∵ ∠CAB = ∠ACD alternate angles)

(iii) ∠ADC = 2 ∠BDC = 2 × 50° = 100°

(∵ ∠ABD = ∠BDC alternate angles)

Question 10.

In the given isosceles trapezium ABCD, ∠C = 102°. Find all the remaining angles of the trapezium.

Solution:

AB || CD

∠B + ∠C = 180°

(∵ adjacent angles on the same side of

transversal are supplementary)

⇒ ∠B + 102° = 180°

∠B = 180°- 102° = 78°

As AD = BC (Given)

∴ ∠A = ∠B = 78°

∠A + ∠B + ∠C + ∠D = 360°

78° + 78° + 102° + ∠D = 360°

∠D + 258° = 360°

∠D = 102°

Question 11.

In the given figure, PQRS is a kite. Find the values of x and y.

Solution:

In the figure, PQRS is a kite

∠Q = 120° and ∠R = 50°

∴ ∠Q = ∠S

∴ x = 120°

∠P + ∠R = 360° – (120° + 120°)

∠P + ∠R = 360° – 240° = 120°

But ∠R = 50°

∴ ∠P = y = 120°- 50° = 70°

Hence, x = 120°, y = 70°