# ML Aggarwal Quadrilaterals Shapes Exe-13.3 Class 8 ICSE Ch-13 Maths Solutions

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 |

**Quadrilaterals Shapes Exe-13.3**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-244

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

(i) four sides of equal length

(ii) four right angles.

**Answer:**

(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.

**Answer:**

(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.

**Answer:**

(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.**

**Answer:**

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.**

**Answer:**

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.**

**Answer:**

ABCD is a trapezium

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

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

(3x + 11)^{o} + (5x – 31)^{o} = 180^{o}

3x + 11 + 5x – 31 = 180^{o}

8x – 20 = 180^{o}

8x = 180^{o} + 20

8x = 200^{o}

x = 200^{o}/8

x = 25^{o}

Then, ∠B = 3x + 11

= (3 × 25) + 11

= 75 + 11

= 86^{o}

∠C = 5x – 31

= (5 × 25) – 31

= 125 – 31

= 94^{o}

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

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

∠A + ∠D = 180^{o}

5y + 7y = 180^{o}

12y = 180^{o}

y = 180^{o}/12

y = 15^{o}

Then, ∠A = 5y = (5 × 15) = 75^{o}

∠D= 7y = (7 × 15) = 105^{o}

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

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

(i) ∠CEB,

(ii) ∠DCF

**Answer:**

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

**Answer:**

ABCD is a rectangle and diagonals intersect at O.

∠AOB = 118^{o}

**(i) Consider the ΔAOB,**

∠OAB = ∠OBA

Let us assume ∠OAB = ∠OBA = y^{o}

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

∠OAB + ∠OBA + ∠AOB = 180^{o}

y + y + 118^{o} = 180^{o}

2x + 118^{o} = 180^{o}

2y = 180^{o} – 118^{o}

2y = 62^{o}

y = 62/2

y = 31^{o}

So, ∠OAB = ∠OBA = 31^{o}

Hence, ∠ABO = 31^{o}

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

∠AOB + ∠AOD = 180^{o}

118^{o} + ∠AOD = 180^{o}

∠AOD = 180^{o} – 118^{o}

∠AOD = 62^{o}

Now consider the ΔAOD,

Let us assume the ∠ADO = ∠DAO = x

∠AOD + ∠ADO + ∠DAO = 180^{o}

62^{o} + x + x = 180^{o}

62^{o} + 2x = 180^{o}

2x = 180^{o} – 62

2x = 118^{o}

x = 118^{o}/2

x = 59^{o}

**Hence, ∠ADO = 59 ^{o}**

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

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

(i) ∠CAB

(ii) ∠BCD

(iii) ∠ADC

**Answer:**

**(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)

**Quadrilaterals Shapes Exe-13.3**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-245

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

**Answer:**

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.**

**Answer:**

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 :–

Return to : **– **ML Aggarwal Maths Solutions for ICSE Class -8

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