## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 13 Understanding Quadrilaterals Ex 13.1

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

Solution:

(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

Solution:

(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

Solution:

(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°

Solution:

(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°

Solution:

(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°

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

As AB || CD

∠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°

Question 11.

From the adjoining figure, find

(i) x

(ii) ∠DAB

(iii) ∠ADB

Solution:

(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 ∆)

Question 12.

Find the angle measure x in the following figures:

Solution:

(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 = \(\frac{280^{\circ}}{2}\) = 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.

(i) In the given figure, find x + y + z.

(ii) In the given figure, find x + y + z + w.

Solution:

(i) In ∆ABC, Sides AB, BC, CA are produce in order in one side.

∠B = 70°, ∠C = 90°

∴ ∠A = 180° – (∠B + ∠C)

= 180°- (70°+ 90°)

= 180°- 160° = 20°

But x + 90° = 180° (Linear pair)

∴ x = 180° – 90° = 90°

Similarly, y = 180° – 70° = 110°

y = 180° – 20°= 160°

x + y + z = 90° + 110°+ 160° = 360°

(ii) In the given figure,

ABCD is a quadrilateral whose sides are produced in order.

∠A = 130°, ∠B = 80°, ∠C = 70°

∴ ∠D = 360° – (∠A + ∠B + ∠C)

= 360° – (130° + 80° + 70°)

= 360° – 280° = 80°

Now, x + 130° = 180° (Linear pair)

∴ x = 180°- 130° = 50°

Similarly, y = 180° – 80° = 100°

z = 180° – 70°= 110°

w = 180°-80°= 100°

∴ x + y + z + w = 52° + 100° +110°+ 100°

= 360°

Question 14.

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

Solution:

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 = \(\frac{540^{\circ}}{4}\) = 135°

Question 15.

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.

Solution:

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°

⇒ \(\frac{180^{\circ}}{6}\) = 30°

(i) Measure of exterior angle x = 30°

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

(iii) Let number of sides = n, then

\(\frac{2 n-4}{n}\) × 90° = 150°

⇒ \(\frac{2 n-4}{n}=\frac{150^{\circ}}{90}=\frac{5}{3}\)

⇒ 6n – 12 = 5n

⇒ 6n – 5n = 12

⇒ n = 12

∴ Number of sides = 12

Question 16.

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

Solution:

In a polygon,

Let exterior angle = x

Then interior angle = 2x

But sum of interior angle and exterior angle = 180°

∴ 2x + x = 180°

⇒ 3x = 180°

⇒ x = \(\frac{180^{\circ}}{3}\) = 60°

∴ Interior angle = 2 × 60° = 120°

Let number of sides of the polygon = x

Then \(\frac{2 n-4}{n}\) × 90° = 120°

⇒ \(\frac{2 n-4}{n}=\frac{120^{\circ}}{90}=\frac{4}{3}\)

⇒ 6n – 12 = 4n

⇒ 6n – 4n = 12

⇒ 2n = 12

⇒ n = \(\frac{12}{2}\) = 6

∴ Number of sides = 6