# ML Aggarwal Quadrilaterals Shapes Exe-13.1 Class 8 ICSE Ch-13 Maths Solutions

ML Aggarwal Quadrilaterals Shapes Exe-13.1 Class 8 ICSE Ch-13 Maths Solutions. We Provide Step by Step Answer of Exe-13.1 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-13.1 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-13.1 Questions |

Edition | 2023-2024 |

**Quadrilaterals Shapes Exe-13.1**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-232

**Question 1. Some figures are given below.**

Classify each of them on the basis of the following:

(a) Simple curve

(b) Simple closed curve

(c) Polygon

(d) Convex polygon

(e) Concave polygon

**Answer:**

(a) (i), (ii), (iii), (v) and (vi) are simple curves.

(b) (iii), (v), (vi) are simple closed curves.

(c) (iii) and (vi) are polygons.

(d) (iii) is a convex polygon.

(e) (v) is a concave polygon.

**Question 2. How many diagonals does each of the following have?**

(a) A convex quadrilateral

(b) A regular hexagon

**Answer:**

(a) A convex quadrilateral: It has two diagonals.

(b) A regular hexagon: It has 9 diagonals as shown.

**Question 3. Find the sum of measures of all interior angles of a polygon with number of sides:**

(i) 8

(ii) 12

**Answer:**

(i) Sum of measures of all interior angles of

8-sided polygon = (2n – 4) × 90°

= (2 × 8 – 4) × 90°

= 12 × 90° = 1080°

(ii) Sum of measures of all interior angles of

12-sided polygon = (2n – 4) × 90°

= (2 × 12 – 4) × 90°

= 18 × 90°= 1800°

**Question 4. Find the number of sides of a regular polygon whose each exterior angles has a measure of**

(i) 24°

(ii) 60°

(iii) 72°

**Answer:**

(i) Let number of sides of the polygon = n

Each exterior angle = 24°

∴ n = 360°/24° = 15 sides

∴ Polygon is of 15 sides.

(ii) Each interior angle of the polygon = 60°

Let number of sides of the polygon = n

∴ n = 360°/60° = 6

∴ Number of sides = 6

(iii) Each interior angle of the polygon = 72°

Let number of sides of the polygon = n

∴ n = 360°/72° = 5

∴ Number of sides = 5

**Question 5. Find the number of sides of a regular polygon if each of its interior angles is**

(i) 90°

(ii) 108°

(iii) 165°

**Answer:**

**(i) The number of sides of a regular polygon whose each interior angles has a measure of 90 ^{o}**

Let us assume the number of sides of the regular polygon be n,

Then, we know that 90^{o} = ((2n – 4)/n) × 90^{o}

90^{o}/90^{o} = (2n – 4)/n

1 = (2n – 4)/n

2n – 4 = n

2n – n = 4

n = 4

Hence, the number of sides of a regular polygon is 4.

**(ii) The number of sides of a regular polygon whose each interior angles has a measure of 108 ^{o}**

Let us assume the number of sides of the regular polygon be n,

Then, we know that 108^{o} = ((2n – 4)/n) × 90^{o}

108^{o}/90^{o} = (2n – 4)/n

6/5 = (2n – 4)/n

5(2n – 4) = 6n

10n – 20 = 6n

10n – 6n = 20

4n = 20

n = 20/4

n = 5

Hence, the number of sides of a regular polygon is 5.

So, it is a Pentagon.

**(iii) The number of sides of a regular polygon whose each interior angles has a measure of 165 ^{o}**

Let us assume the number of sides of the regular polygon be n,

Then, we know that 165^{o} = ((2n – 4)/n) × 90^{o}

165^{o}/90^{o} = (2n – 4)/n

11/6 = (2n – 4)/n

6(2n – 4) = 11n

12n – 24 = 11n

12n – 11n = 24

n = 24

Hence, the number of sides of a regular polygon is 24.

**Question 6. Find the number of sides in a polygon if the sum of its interior angles is:**

(i) 1260°

(ii) 1980°

(iii) 3420°

**Answer:**

We know that, sum of interior angles of polygon

is given by (2n – 4) at right angles.

**(i) 1260°**

∴ 1260 = (2n – 4) × 90

⇒ 1260/90 = 2n – 4

⇒ 14 = 2n – 4

⇒ n = 9

**(ii) 1980°**

∴ 1980 = (2n – 4) × 90

⇒ 1980/90 = 2n – 4

⇒ 22 = 2n – 4

⇒ n = 13.

**(iii) 3420°**

∴ 3420 = (272 – 4) × 90 3420

⇒ 3420/90 = 2n – 4

⇒ 38 = 2n – 4

⇒ n = 21

**Question 7. If the angles of a pentagon are in the ratio 7 : 8 : 11 : 13 : 15, find the angles.**

**Answer:**

Ratio in the angles of a polygon = 7 : 8 : 11 : 13 : 15

Sum of angles of a pentagon = (2n – 4) × 90°

= (2 × 5 – 4) × 90°

= 6 × 90° = 540°

Let the angles of the pentagon be

7x, 8x, 11x, 13x, 15x

∴ 7x + 8x + 11x + 13x + 15x = 540°

⇒ 54x = 540° ⇒ x = 540°/54 = 10°

∴ Angles are 7 × 10° = 70°, 8 × 10° = 80°,

11 × 10° = 110°, 13 × 10° = 130° and 15 × 10°= 150°

∴ Angles are 70°, 80°, 110°, 130° and 150°

**Question 8. The angles of a pentagon are x°, (x – 10)°, (x + 20)°, (2x – 44)° and (2x – 70°) Calculate x.**

**Answer:**

Angles of a pentaon are x°, (x – 10)°, (x + 20)°,

(2x – 44)° and (2x – 70°)

But sum of angles of a pentagon

= (2n – 4) × 90°

= (2 × 5 – 4) × 90°

= 6 × 90° = 540°

∴ x + x – 10° + x + 20° + 2x – 44° + 2x – 70° = 540°

⇒ 7x – 104° = 540°

⇒ 7x = 540° + 104° = 644°

⇒ x = 644°/7 = 92°

∴ x = 92°

**(ML Aggarwal Quadrilaterals Shapes Exe-13.1 Class 8 ICSE Maths)**

**Question 9. The exterior angles of a pentagon are in ratio 1 : 2 : 3 : 4 : 5. Find all the interior angles of the pentagon.**

**Answer:**

Let the exterior angles of the pentagon are x, 2x, 3x, 4x and 5x.

We know that sum of exterior angles of polygon is 360°.

∴ x + 2x + 3x + 4x + 5x = 360°

⇒ 15x = 360°

⇒ x = 360°/15

⇒ x = 24°

∴ Exterior angles are 24°, 48°, 72°, 96°, 120°

Interior angles are 180° – 24°, 180° – 48°, 180° – 72°, 180° – 96°,

180° – 120° i.e. 156°, 132°, 108°, 84°, 60°.

**Question 10. In a quadrilateral ABCD, AB || DC. If ∠A : ∠D = 2:3 and ∠B : ∠C = ∠7 : 8, find the measure of each angle.**

**Answer:**

In a quadrilateral ABCD, AB || DC. If ∠A : ∠D = 2:3 and ∠B : ∠C = ∠7 : 8,

∠A + ∠D = 180^{o}

Let us assume the angle ∠A = 2a and ∠D = 3a

2a + 3a = 180^{o}

5a = 180^{o}

a = 180^{o}/5

a = 36^{o}

∠A = 2a = 2 × 36^{o} = 72^{o}

∠D = 3a = 3 × 36^{o} = 108^{o}

Now, ∠B + ∠C = 180^{o}

Let us assume the angle ∠B = 7b and ∠C = 8b

7b + 8b = 180^{o}

15b = 180^{o}

b = 180^{o}/15

b = 12^{o}

∠B = 7b = 7 × 12^{o} = 84^{o}

**∠C = 8b = 8 × 12 ^{o} = 96^{o}**

**Question 11. From the adjoining figure, find**

(i) x

(ii) ∠DAB

(iii) ∠ADB

**Answer:**

**(i) ABCD is a quadrilateral**

∴ ∠A + ∠B + ∠C + ∠D = 360°

⇒ (3x + 4) + (50 + x) + (5x + 8) + (3x + 10) = 360

⇒ 3x + 4 + 50 + x + 5x + 8 + 3x + 10 = 360°

⇒ 12x + 72 = 360°

⇒ 12x = 288

⇒ x = 24

(ii) ∠DAB = (3x + 4) = 3 × 24 + 4 = 76°

(iii) ∠ADB = 180°- (76° + 50°) = 54° (∵ ABD is a ∆)

**Quadrilaterals Shapes Exe-13.1**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-233

**Question 12. Find the angle measure x in the following figures:**

**Answer:**

**(i) In quadrilateral three angles are 40°, 140° and 100°**

But sum of Four angles = 360°

⇒ 40° + 140°+ 100° + x = 360°

⇒ 280° + x = 360°

⇒ x = 360° – 280° = 80°

**(ii) In the given figure, ABCDE is a pentagon.**

Where side AB is produced to both sides0

∠1 + 60° = 180° (Linear pair)

∠1 = 180°- 60°= 120°

Similarly ∠2 + 80° = 180°

∴ ∠2 = 180°- 80°= 100°

Now, sum of angles of a pentagon = (2n – 4) × 90°

= (2 × 5 – 4) × 90° = 6 × 90° = 540°

∴ ∠A + ∠B + ∠C + ∠D + ∠E = 540°

⇒ 120° + 100° + x + 40° + x = 540°

⇒ 260° + 2x = 540°

⇒ 2x = 540° – 260° = 280°

⇒ x = 280°/2 = 140°

**(iii) In the given figure, ABCD is a quadrilateral**

whose side AB is produced is both sides ∠A = 90°

But ∠A + ∠B + ∠C + ∠D = 360°

(Sum of angles of a quadrilateral)

⇒ 90°+ 60°+ 110° + x = 360°

⇒ 260° + x = 360°

⇒ x = 360° – 260° = 100°

∴ x = 100°

**(iv) In the given figure, ABCD is a quadrilateral**

whose side AB is produced to E.

∠A = 90°, ∠C = 83°, ∠D = 110°

∠B + x = 180° (Lienar pair)

∠B = 180° – x

But ∠A + ∠B + ∠C + ∠D = 360°

⇒ 90° + (180° – x) + 83° + 110° = 360°

(Sum of angles of a quadrilateral)

⇒ 283°+ 180° – x = 360°

⇒ x = 283° + 180°- 360°

⇒ x = 463° – 360°= 103°

**Question 13. A heptagon has three equal angles each of 120° and four equal angles. Find the size of equal angles.**

**Answer:**

Sum of angles of a heptagon = (2 × n – 4) × 90°

= (2 × 7 – 4) × 90°

= 10 × 90° = 900°

Sum of three angles are each equal i.e. 120°

= 120° × 3 = 360°

Sum of remaining 4 equal angles

= 900° – 360° = 540°

∴ Each angle = 540°/4 = 135°

**Question 14. The ratio between an exterior angle and the interior angle of a regular polygon is 1 : 5. Find**

(i) the measure of each exterior angle

(ii) the measure of each interior angle

(iii) the number of sides in the polygon.

**Answer:**

Ratio between an exterior and an interior angle = 1 : 5

Let exterior angle = x

Then interior angle = 5x

But sum of interior angle and exterior angle = 180°

∴ x + 5x = 180°

⇒ 6x = 180°

⇒ 180°/6 = 30°

(i) Measure of exterior angle x = 30°

(ii) and measure of interior angle = 5x = 5 × 30° = 150°

(iii) the number of sides in the polygon

The number of sides of a regular polygon whose each interior angles has a measure of 150^{o}

Let us assume the number of sides of the regular polygon be n,

150^{o} = ((2n – 4)/n) × 90^{o}

150^{o}/90^{o} = (2n – 4)/n

5/3 = (2n – 4)/n

By cross multiplication,

3(2n – 4) = 5n

6n – 12 = 5n

6n – 5n = 12

n = 12

Therefore, the number of sides of a regular polygon is 12.

**(ML Aggarwal Quadrilaterals Shapes Exe-13.1 Class 8 ICSE Maths)**

**Question 15. Each interior angle of a regular polygon is double of its exterior angle. Find the number of sides in the polygon.**

**Answer:**

Each interior angle of a regular polygon is double of its exterior angle.

So, let us assume exterior angle be y

Interior angle be 2y,

sum of interior and exterior angle is equal to 180^{o},

y + 2y = 180^{o}

3y = 180^{o}

y = 180^{o}/3

y = 60^{o}

interior angle = 2y = 2 × 60^{o} = 120^{o}

The number of sides of a regular polygon whose each interior angles has a measure of 120^{o}

Let us assume the number of sides of the regular polygon be n,

Then, we know that 120^{o} = ((2n – 4)/n) × 90^{o}

120^{o}/90^{o} = (2n – 4)/n

4/3 = (2n – 4)/n

By cross multiplication,

3(2n – 4) = 4n

6n – 12 = 4n

6n – 4n = 12

2n = 12

n = 12/2

n = 6

Hence, the number of sides of a regular polygon is 6.

— End of Quadrilaterals Shapes **Exe-13.1 **Class 8 ICSE Maths Solutions :–

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

Thanks