ML Aggarwal Quadrilaterals Shapes Exe-13.3 Class 8 ICSE Ch-13 Maths Solutions. We Provide Step by Step Answer of  Exe-13.3 Questions for Quadrilaterals Shapes as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

## ML Aggarwal Quadrilaterals Shapes Exe-13.3 Class 8 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 8th Chapter-13 Quadrilaterals Shapes Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-13.3 Questions Edition 2023-2024

ML Aggarwal Class 8 ICSE Maths Solutions

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#### Question 1. Identify all the quadrilaterals that have

(i) four sides of equal length
(ii) four right angles.

(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

(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle.

(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.

(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.

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.

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,
AB2 = AO2 + BO2
AB2 = 82 + 62
AB2 = 64 + 36
AB2 = 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.

ABCD is a trapezium

∠A : ∠D = 5 : 7, ∠B = (3x + 11)o and ZC = (5x – 31)o

Then, ∠B + ∠C = 180o … [because co – interior angle]

(3x + 11)o + (5x – 31)o = 180o

3x + 11 + 5x – 31 = 180o

8x – 20 = 180o

8x = 180o + 20

8x = 200o

x = 200o/8

x = 25o

Then, ∠B = 3x + 11

= (3 × 25) + 11

= 75 + 11

= 86o

∠C = 5x – 31

= (5 × 25) – 31

= 125 – 31

= 94o

let us assume the angles ∠A = 5y and ∠D = 7y

sum of co – interior angles are equal to 180o.

∠A + ∠D = 180o

5y + 7y = 180o

12y = 180o

y = 180o/12

y = 15o

Then, ∠A = 5y = (5 × 15) = 75o

∠D= 7y = (7 × 15) = 105o

Therefore, the angles are ∠A = 75o, ∠B = 86o, ∠C = 94o and ∠D = 105o.

#### Question 7. In the given figure, ABCD is a rectangle. If ∠CEB : ∠ECB = 3 : 2 find

(i) ∠CEB,
(ii) ∠DCF

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
(iii) ∠OCB

ABCD is a rectangle and diagonals intersect at O.

∠AOB = 118o

(i) Consider the ΔAOB,

∠OAB = ∠OBA

Let us assume ∠OAB = ∠OBA = yo

Sum of measures of interior angles of triangle is equal to 180o.

∠OAB + ∠OBA + ∠AOB = 180o

y + y + 118o = 180o

2x + 118o = 180o

2y = 180o – 118o

2y = 62o

y = 62/2

y = 31o

So, ∠OAB = ∠OBA = 31o

Hence, ∠ABO = 31o

(ii) We know that sum of liner pair angles is equal to 180o.

∠AOB + ∠AOD = 180o

118o + ∠AOD = 180o

∠AOD = 180o – 118o

∠AOD = 62o

Now consider the ΔAOD,

Let us assume the ∠ADO = ∠DAO = x

∠AOD + ∠ADO + ∠DAO = 180o

62o + x + x = 180o

62o + 2x = 180o

2x = 180o – 62

2x = 118o

x = 118o/2

x = 59o

(iii) ∠OCB = ∠OAD = 59o  … [because alternate angles are equal]

#### Question 9. In the given figure, ABCD is a rhombus and ∠ABD = 50°. Find :

(i) ∠CAB
(ii) ∠BCD

(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)

ML Aggarwal Class 8 ICSE Maths Solutions

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#### Question 10. In the given isosceles trapezium ABCD, ∠C = 102°. Find all the remaining angles of the trapezium.

AB || CD
∠B + ∠C = 180°
(∵ adjacent angles on the same side of
transversal are supplementary)
⇒ ∠B + 102° = 180°
∠B = 180°- 102° = 78°
∴ ∠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.

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°

— End of Quadrilaterals Shapes Exe-13.3 Class 8 ICSE Maths Solutions :–