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Understanding Quadrilaterals Class-8 ML Aggarwal ICSE Class-8 Maths

Understanding Quadrilaterals Class-8 ML Aggarwal ICSE Mathematics Solutions Chapter-13. We provide step by step Solutions of Exercise / lesson-13 Understanding Quadrilaterals Class-8th ML Aggarwal ICSE Mathematics.

Our Solutions contain all type Questions with Exe-13.1 , Exe-13.2, Exe-13.3 , Objective Type Questions (including Mental Maths Multiple Choice Questions, HOTS,) and Check Your Progress to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics .

Understanding Quadrilaterals Class-8 ML Aggarwal ICSE Mathematics Solutions Chapter-13


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 Exercise 13.1 ,

Exercise-13.2,

Exercise-13.3,

Objective Type Questions, 

Mental Maths,

Multiple Choice Questions (MCQ)

HOTS

Check Your Progress


Exe-13.1 Understanding Quadrilaterals Class-8 ML Aggarwal ICSE  Maths Solutions

Question 1.
Some figures are given below.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 1
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

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(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.
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 2

Question 3.
Find the sum of measures of all interior angles of a polygon with number of sides:
(i) 8
(ii) 12

Answer

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(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 = \frac{360^{\circ}}{24^{\circ}} = 15 sides
∴ Polygon is of 15 sides.
(ii) Each interior angle of the polygon = 60°
Let number of sides of the polygon = n
∴ n = \frac{360^{\circ}}{60^{\circ}} = 6
∴ Number of sides = 6
(iii) Each interior angle of the polygon = 72°
Let number of sides of the polygon = n
∴ n = \frac{360^{\circ}}{72^{\circ}} = 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) Each interior angle = 90°
Let number of sides of the regular polgyon = n
∴ 90° = \frac{2 n-4}{n} × 90°
⇒ \frac{2 n-4}{n}=\frac{90^{\circ}}{90^{\circ}} = 1
⇒ 2n – 4 = n
⇒ 2n – n = 4
⇒ n = 4
⇒ n = 4
∴ It is a square.
(ii) Each interior angle = 108°
Let number of sides of the regular polygon = n
∴ 108° = \frac{2 n-4}{n} × 90°
⇒ \frac{2 n-4}{n}=\frac{108^{\circ}}{90^{\circ}}=\frac{6}{5}
⇒ 10n – 20 = 6n ⇒ 10n – 6n = 20
⇒ 4n = 20
⇒ n = \frac{20}{4} = 5
∴ It is a pentagon.
(iii) Each interior angle = 165°
Let number of sides of the regular polygon = n
∴ 165° = \frac{2 n-4}{n} × 90°
⇒ \frac{2 n-4}{n}=\frac{165^{\circ}}{90^{\circ}}=\frac{11}{6}
⇒ 12n – 24 = 11n
⇒ 12n – 11n = 24
⇒ n = 24
∴ It is 24-sided polygon.

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

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We know that, sum of interior angles of polygon
is given by (2n – 4) at right angles.
(i) 1260°
∴ 1260 = (2n – 4) × 90
⇒ \frac{1260}{90} = 2n – 4
⇒ 14 = 2n – 4
⇒ n = 9
(ii) 1980°
∴ 1980 = (2n – 4) × 90
⇒ \frac{1980}{90} = 2n – 4
⇒ 22 = 2n – 4
⇒ n = 13.
(iii) 3420°
∴ 3420 = (272 – 4) × 90 3420
⇒ \frac{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 = \frac{540^{\circ}}{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 = \frac{644^{\circ}}{7} = 92°
∴ x = 92°

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 = \frac{360^{\circ}}{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

As AB || CD
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1 3
∠A + ∠D = 180° and ∠B + ∠C = 180°
⇒ 2x + 3x = 180° and 7y + 8y = 180°
5x = 180° and 15y = 180°
x = 36° and y = 12°
∴ ∠A = 2 × 36 = 72°
and ∠D = 3 × 36 = 108°
∠B = 7y = 7 × 12 = 84°
and ∠C = 8y = 8 × 12 = 96°

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