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

ML Aggarwal Class 8 ICSE Maths Solutions

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#### 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

(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?

(b) A regular hexagon

(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

(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°

(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°

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

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

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

90o/90o = (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 108o

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

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

108o/90o = (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 165o

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

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

165o/90o = (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°

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.

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.

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.

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.

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

∠A + ∠D = 180o

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

2a + 3a = 180o

5a = 180o

a = 180o/5

a = 36o

∠A = 2a = 2 × 36o = 72o

∠D = 3a = 3 × 36o = 108o

Now, ∠B + ∠C = 180o

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

7b + 8b = 180o

15b = 180o

b = 180o/15

b = 12o

∠B = 7b = 7 × 12o = 84o

∠C = 8b = 8 × 12o = 96o

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

(i) x
(ii) ∠DAB

∴ ∠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 ∆)

ML Aggarwal Class 8 ICSE Maths Solutions

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

(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.

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.

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 150o

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

150o = ((2n – 4)/n) × 90o

150o/90o = (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.

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 180o,

y + 2y = 180o

3y = 180o

y = 180o/3

y = 60o

interior angle = 2y = 2 × 60o = 120o

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

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

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

120o/90o = (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 :–