## Selina Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes (Including Polygons)

**Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 16 Understanding Shapes (Including Polygons)**

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### Understanding Shapes Exercise 16A – Selina Concise Mathematics Class 8 ICSE Solutions

**Question 1.**

State which of the following are polygons :

If the given figure is a polygon, name it as convex or concave.

**Solution:**

Only Fig. (ii), (iii) and (v) are polygons.

Fig. (ii) and (iii) are concave polygons while

Fig. (v) is convex.

**Question 2.**

Calculate the sum of angles of a polygon with :

(i) 10 sides

(ii) 12 sides

(iii) 20 sides

(iv) 25 sides

**Solution:**

(i) No. of sides n = 10

sum of angles of polygon = (n – 2) x 180°

= (10 – 2) x 180° = 1440°

(ii) no. of sides n = 12

sum of angles = (n – 2) x 180°

= (12 – 2) x 180° = 10 x 180° = 1800°

(iii) n = 20

Sum of angles of Polygon = (n – 2) x 180°

= (20 – 2) x 180° = 3240°

(iv) n = 25

Sum of angles of polygon = (n – 2) x 180°

= (25 – 2) x 180° = 4140°

**Question 3.**

Find the number of sides in a polygon if the sum of its interior angles is :

(i) 900°

(ii) 1620°

(iii) 16 right-angles

(iv) 32 right-angles.

**Solution:**

(i) Let no. of sides = n

Sum of angles of polygon = 900°

(n – 2) x 180° = 900°

n – 2 = \(\frac { 900 }{ 180 }\)

n – 2 = 5

n = 5 + 2

n = 7

(ii) Let no. of sides = n

Sum of angles of polygon = 1620°

(n – 2) x 180° = 1620°

n – 2 = \(\frac { 1620 }{ 180 }\)

n – 2 = 9

n = 9 + 2

n = 11

(iii) Let no. of sides = n

Sum of angles of polygon = 16 right angles = 16 x 90 = 1440°

(n – 2) x 180° = 1440°

n – 2 = \(\frac { 1440 }{ 180 }\)

n – 2 = 8

n = 8 + 2

n = 10

(iv) Let no. of sides = n

Sum of angles of polygon = 32 right angles = 32 x 90 = 2880°

(n – 2) x 180° = 2880

n – 2 = \(\frac { 2880 }{ 180 }\)

n – 2 = 16

n = 16 + 2

n = 18

**Question 4.**

Is it possible to have a polygon ; whose sum of interior angles is :

(i) 870°

(ii) 2340°

(iii) 7 right-angles

(iv) 4500°

**Solution:**

(i) Let no. of sides = n

Sum of angles = 870°

(n – 2) x 180° = 870°

n – 2 = \(\frac { 870 }{ 180 }\)

n – 2 = \(\frac { 29 }{ 6 }\)

n = \(\frac { 29 }{ 6 }\) + 2

n = \(\frac { 41 }{ 6 }\)

Which is not a whole number.

Hence it is not possible to have a polygon, the sum of whose interior angles is 870°

(ii) Let no. of sides = n

Sum of angles = 2340°

(n – 2) x 180° = 2340°

n – 2 = \(\frac { 2340 }{ 180 }\)

n – 2 = 13

n = 13 + 2 = 15

Which is a whole number.

Hence it is possible to have a polygon, the sum of whose interior angles is 2340°.

(iii) Let no. of sides = n

Sum of angles = 7 right angles = 7 x 90 = 630°

(n – 2) x 180° = 630°

n – 2 = \(\frac { 630 }{ 180 }\)

n – 2 = \(\frac { 7 }{ 2 }\)

n = \(\frac { 7 }{ 2 }\) + 2

n = \(\frac { 11 }{ 2 }\)

Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is 7 right-angles.

(iv) Let no. of sides = n

(n – 2) x 180° = 4500°

n – 2 = \(\frac { 4500 }{ 180 }\)

n – 2 = 25

n = 25 + 2

n = 27

Which is a whole number.

Hence it is possible to have a polygon, the sum of whose interior angles is 4500°.

**Question 5.**

(i) If all the angles of a hexagon are equal ; find the measure of each angle.

(ii) If all the angles of a 14-sided figure are equal ; find the measure of each angle.

**Solution:**

(i) No. of sides of hexagon, n = 6

Let each angle be = x°

Sum of angles = 6x°

(n – 2) x 180° = Sum of angles

(6 – 2) x 180° = 6x°

4 x 180 = 6x

**Question 6.**

Find the sum of exterior angles obtained on producing, in order, the sides of a polygon with :

(i) 7 sides

(ii) 10 sides

(iii) 250 sides.

**Solution:**

(i) No. of sides n = 7

Sum of interior & exterior angles at one vertex = 180°

**Question 7.**

The sides of a hexagon are produced in order. If the measures of exterior angles so obtained are (6x – 1)°, (10x + 2)°, (8x + 2)° (9x – 3)°, (5x + 4)° and (12x + 6)° ; find each exterior angle.

**Solution:**

Sum of exterior angles of hexagon formed by producing sides of order = 360°

i.e. 41° ; 72°, 58° ; 60° ; 39° and 90°

**Question 8.**

The interior angles of a pentagon are in the ratio 4 : 5 : 6 : 7 : 5. Find each angle of the pentagon.

**Solution:**

Let the interior angles of the pentagon be 4x, 5x, 6x, 7x, 5x.

Their sum = 4x + 5x + 6x + 7x + 5x = 21x

**Question 9.**

Two angles of a hexagon are 120° and 160°. If the remaining four angles are equal, find each equal angle.

**Solution:**

Two angles of a hexagon are 120°, 160°

Let remaining four angles be x, x, x and x.

Their sum = 4x + 280°

But sum of all the interior angles of a hexagon

**Question 10.**

The figure, given below, shows a pentagon ABCDE with sides AB and ED parallel to each other, and ∠B : ∠C : ∠D = 5 : 6 : 7.

(i) Using formula, find the sum of interior angles of the pentagon.

(ii) Write the value of ∠A + ∠E

(iii) Find angles B, C and D.

**Solution:**

(i) Sum of interior angles of the pentagon

**Question 11.**

Two angles of a polygon are right angles and the remaining are 120° each. Find the number of sides in it.

**Solution:**

Let number of sides = n

n = \(\frac { 300 }{ 60 }\)

n = 5

**Question 12.**

In a hexagon ABCDEF, side AB is parallel to side FE and ∠B : ∠C : ∠D : ∠E = 6 : 4 : 2 : 3. Find ∠B and ∠D.

**Solution:**

**Question 13.**

the angles of a hexagon are x + 10°, 2x + 20°, 2x – 20°, 3x – 50°, x + 40° and x + 20°. Find x.

**Solution:**

**Question 14.**

In a pentagon, two angles are 40° and 60°, and the rest are in the ratio 1 : 3 : 7. Find the biggest angle of the pentagon.

**Solution:**

In a pentagon, two angles are 40° and 60° Sum of remaining 3 angles = 3 x 180°

= 540° – 40° – 60° = 540° – 100° = 440°

Ratio in these 3 angles =1 : 3 : 7

Sum of ratios =1 + 3 + 7 = 11

Biggest angle = \(\frac { 440\times 7 }{ 11 }\) = 280°

### Understanding Shapes Exercise 16B – Selina Concise Mathematics Class 8 ICSE Solutions

**Question 1.**

Fill in the blanks :

In case of regular polygon, with :

**Solution:**

**Question 2.**

Find the number of sides in a regular polygon, if its each interior angle is :

(i) 160°

(ii) 135°

(iii) \(1\frac { 1 }{ 5 }\) of a right-angle

**Solution:**

**Question 3.**

Find the number of sides in a regular polygon, if its each exterior angle is :

(i) \(\frac { 1 }{ 3 }\) of a right angle

(ii) two-fifth of a right-angle.

**Solution:**

(i) Each exterior angle = \(\frac { 1 }{ 3 }\) of a right angle

= \(\frac { 1 }{ 3 }\) x 90

= 30°

Let number of sides = n

**Question 4.**

Is it possible to have a regular polygon whose each interior angle is :

(i) 170°

(ii) 138°

**Solution:**

(i) No. of sides = n

each interior angle = 170°

Which is not a whole number.

Hence it is not possible to have a regular polygon having interior angle of 138°.

**Question 5.**

Is it possible to have a regular polygon whose each exterior angle is :

(i) 80°

(ii) 40% of a right angle.

**Solution:**

(i) Let no. of sides = n each exterior angle = 80°

Which is not a whole number.

Hence it is not possible to have a regular polygon whose each exterior angle is of 80°.

(ii) Let number of sides = n

Each exterior angle = 40% of a right angle

Which is a whole number.

Hence it is possible to have a regular polygon whose each exterior angle is 40% of a right angle.

**Question 6.**

Find the number of sides in a regular polygon, if its interior angle is equal to its exterior angle.

**Solution:**

Let each exterior angle or interior angle be = x°

**Question 7.**

The exterior angle of a regular polygon is one-third of its interior angle. Find the number of sides in the polygon.

**Solution:**

Let interior angle = x°

Exterior angle = \(\frac { 1 }{ 3 }\) x°

**Question 8.**

The measure of each interior angle of a regular polygon is five times the measure of its exterior angle. Find :

(i) measure of each interior angle ;

(ii) measure of each exterior angle and

(iii) number of sides in the polygon.

**Solution:**

Let exterior angle = x°

Interior angle = 5x°

x + 5x = 180°

6x = 180°

x = 30°

Each exterior angle = 30°

Each interior angle = 5 x 30° = 150°

Let no. of sides = n

**Question 9.**

The ratio between the interior angle and the exterior angle of a regular polygon is 2 : 1. Find :

(i) each exterior angle of the polygon ;

(ii) number of sides in the polygon

**Solution:**

Interior angle : exterior angle = 2 : 1

Let interior angle = 2x° & exterior angle = x°

**Question 10.**

The ratio between the exterior angle and the interior angle of a regular polygon is 1 : 4. Find the number of sides in the polygon.

**Solution:**

Let exterior angle = x° & interior angle = 4x°

**Question 11.**

The sum of interior angles of a regular polygon is twice the sum of its exterior angles. Find the number of sides of the polygon.

**Solution:**

Let number of sides = n

Sum of exterior angles = 360°

Sum of interior angles = 360° x 2 = 720°

Sum of interior angles = (n – 2) x 180°

720° = (n – 2) x 180°

n – 2 = \(\frac { 720 }{ 180 }\)

n – 2 = 4

n = 4 + 2

n = 6

**Question 12.**

AB, BC and CD are three consecutive sides of a regular polygon. If angle BAC = 20° ; find :

(i) its each interior angle,

(ii) its each exterior angle

(iii) the number of sides in the polygon.

**Solution:**

**Question 13.**

Two alternate sides of a regular polygon, when produced, meet at the right angle. Calculate the number of sides in the polygon.

**Solution:**

**Question 14.**

In a regular pentagon ABCDE, draw a diagonal BE and then find the measure of:

(i) ∠BAE

(ii) ∠ABE

(iii) ∠BED

**Solution:**

(i) Since number of sides in the pentagon = 5

Each exterior angle = \(\frac { 360 }{ 5 }\) = 72°

∠BAE = 180° – 72°= 108°

**Question 15.**

The difference between the exterior angles of two regular polygons, having the sides equal to (n – 1) and (n + 1) is 9°. Find the value of n.

**Solution:**

We know that sum of exterior angles of a polynomial is 360°

(i) If sides of a regular polygon = n – 1

**Question 16.**

If the difference between the exterior angle of a n sided regular polygon and an (n + 1) sided regular polygon is 12°, find the value of n.

**Solution:**

We know that sum of exterior angles of a polygon = 360°

Each exterior angle of a regular polygon of 360°

**Question 17.**

The ratio between the number of sides of two regular polygons is 3 : 4 and the ratio between the sum of their interior angles is 2 : 3. Find the number of sides in each polygon.

**Solution:**

Ratio of sides of two regular polygons = 3 : 4

Let sides of first polygon = 3n

and sides of second polygon = 4n

Sum of interior angles of first polygon

**Question 18.**

Three of the exterior angles of a hexagon are 40°, 51 ° and 86°. If each of the remaining exterior angles is x°, find the value of x.

**Solution:**

Sum of exterior angles of a hexagon = 4 x 90° = 360°

Three angles are 40°, 51° and 86°

Sum of three angle = 40° + 51° + 86° = 177°

Sum of other three angles = 360° – 177° = 183°

Each angle is x°

3x = 183°

x = \(\frac { 183 }{ 3 }\)

Hence x = 61

**Question 19.**

Calculate the number of sides of a regular polygon, if:

(i) its interior angle is five times its exterior angle.

(ii) the ratio between its exterior angle and interior angle is 2 : 7.

(iii) its exterior angle exceeds its interior angle by 60°.

**Solution:**

Let number of sides of a regular polygon = n

(i) Let exterior angle = x

Then interior angle = 5x

x + 5x = 180°

=> 6x = 180°

**Question 20.**

The sum of interior angles of a regular polygon is thrice the sum of its exterior angles. Find the number of sides in the polygon.

**Solution:**

Sum of interior angles = 3 x Sum of exterior angles

Let exterior angle = x

The interior angle = 3x

x + 3x=180°

=> 4x = 180°

=> x = \(\frac { 180 }{ 4 }\)

=> x = 45°

Number of sides = \(\frac { 360 }{ 45 }\) = 8

### Understanding Shapes Exercise 16C – Selina Concise Mathematics Class 8 ICSE Solutions

**Question 1.**

Two angles of a quadrilateral are 89° and 113°. If the other two angles are equal; find the equal angles.

**Solution:**

Let the other angle = x°

According to given,

89° + 113° + x° + x° = 360°

2x° = 360° – 202°

2x° = 158°

x° = \(\frac { 158 }{ 2 }\)

other two angles = 79° each

**Question 2.**

Two angles of a quadrilateral are 68° and 76°. If the other two angles are in the ratio 5 : 7; find the measure of each of them.

**Solution:**

Two angles are 68° and 76°

Let other two angles be 5x and 7x

68° + 76°+ 5x + 7x = 360°

12x + 144° = 360°

12x = 360° – 144°

12x = 216°

x = 18°

angles are 5x and 7x

i.e. 5 x 18° and 7 x 18° i.e. 90° and 126°

**Question 3.**

Angles of a quadrilateral are (4x)°, 5(x+2)°, (7x – 20)° and 6(x+3)°. Find :

(i) the value of x.

(ii) each angle of the quadrilateral.

**Solution:**

Angles of quadrilateral are,

**Question 4.**

Use the information given in the following figure to find :

(i) x

(ii) ∠B and ∠C

**Solution:**

**Question 5.**

In quadrilateral ABCD, side AB is parallel to side DC. If ∠A : ∠D = 1 : 2 and ∠C : ∠B = 4 : 5

(i) Calculate each angle of the quadrilateral.

(ii) Assign a special name to quadrilateral ABCD

**Solution:**

**Question 6.**

From the following figure find ;

(i) x

(ii) ∠ABC

(iii) ∠ACD

**Solution:**

(i) In Quadrilateral ABCD,

x + 4x + 3x + 4x + 48° = 360°

12x = 360° – 48°

12x = 312

**Question 7.**

Given : In quadrilateral ABCD ; ∠C = 64°, ∠D = ∠C – 8° ; ∠A = 5(a+2)° and ∠B = 2(2a+7)°.

Calculate ∠A.

**Solution:**

∠C = 64° (Given)

∠D = ∠C – 8° = 64°- 8° = 56°

∠A = 5(a+2)°

∠B = 2(2a+7)°

Now ∠A + ∠B + ∠C + ∠D = 360°

5(a+2)° + 2(2a+7)° + 64° + 56° = 360°

5a + 10 + 4a + 14° + 64° + 56° = 360°

9a + 144° = 360°

9a = 360° – 144°

9a = 216°

a = 24°

∠A = 5 (a + 2) = 5(24+2) = 130°

**Question 8.**

In the given figure : ∠b = 2a + 15 and ∠c = 3a + 5; find the values of b and c.

**Solution:**

Stun of angles of quadrilateral = 360°

70° + a + 2a + 15 + 3a + 5 = 360°

6a + 90° = 360°

6a = 270°

a = 45°

b = 2a + 15 = 2 x 45 + 15 = 105°

c = 3a + 5 = 3 x 45 + 5 = 140°

Hence ∠b and ∠c are 105° and 140°

**Question 9.**

Three angles of a quadrilateral are equal. If the fourth angle is 69°; find the measure of equal angles.

**Solution:**

Let each equal angle be x°

x + x + x + 69° = 360°

3x = 360°- 69

3x = 291

x = 97°

Each, equal angle = 97°

**Question 10.**

In quadrilateral PQRS, ∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7.

Calculate each angle of the quadrilateral and then prove that PQ and SR are parallel to each other

(i) Is PS also parallel to QR ?

(ii) Assign a special name to quadrilateral PQRS.

**Solution:**

**Question 11.**

Use the informations given in the following figure to find the value of x.

**Solution:**

Take A, B, C, D as the vertices of Quadrilateral and BA is produced to E (say).

Since ∠EAD = 70°

∠DAB = 180° – 70°= 110°

[EAB is a straight line and AD stands on it ∠EAD+ ∠DAB = 180°]

110° + 80° + 56° + 3x – 6° = 360°

[sum of interior angles of a quadrilateral = 360°]

3x = 360° – 110° – 80° – 56° + 6°

3x = 360° – 240° = 120°

x = 40°

**Question 12.**

The following figure shows a quadrilateral in which sides AB and DC are parallel. If ∠A : ∠D = 4 : 5, ∠B = (3x – 15)° and ∠C = (4x + 20)°, find each angle of the quadrilateral ABCD.

**Solution:**

Let ∠A = 4x

∠D = 5x

Since ∠A + ∠D = 180° [AB||DC]

4x + 5x = 180°

=> 9x = 180°

=> x = 20°

∠A = 4 (20) = 80°,

∠D = 5 (20) = 100°

Again ∠B + ∠C = 180° [ AB||DC]

3x – 15° + 4x + 20° = 180°

7x = 180° – 5°

=> 7x = 175°

=> x = 25°

∠B = 75° – 15° = 60°

and ∠C = 4 (25) + 20 = 100°+ 20°= 120°

**Question 13.**

Use the following figure to find the value of x

**Solution:**

The sum of exterior angles of a quadrilateral

=> y + 80° + 60° + 90° = 360°

=> y + 230° = 360°

=> y = 360° – 230° = 130°

At vertex A,

∠y + ∠x = 180° (Linear pair)

x = 180° – 130°

=> x = 50°

**Question 14.**

ABCDE is a regular pentagon. The bisector of angle A of the pentagon meets the side CD in point M. Show that ∠AMC = 90°.

**Solution:**

Given : ABCDE is a regular pentagon.

The bisector ∠A of the pentagon meets the side CD at point M.

To prove : ∠AMC = 90°

Proof: We know that, the measure of each interior angle of a regular pentagon is 108°.

∠BAM = \(\frac { 1 }{ 2 }\) x 108° = 54°

Since, we know that the sum of a quadrilateral is 360°

In quadrilateral ABCM, we have

∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°

54° + 108° + 108° + ∠AMC = 360°

∠AMC = 360° – 270°

∠AMC = 90°

**Question 15.**

In a quadrilateral ABCD, AO and BO are bisectors of angle A and angle B respectively. Show that:

∠AOB = \(\frac { 1 }{ 2 }\) (∠C + ∠D)

**Solution:**

Given : AO and BO are the bisectors of ∠A and ∠B respectively.

∠1 = ∠4 and ∠3 = ∠5 ……..(i)

To prove : ∠AOB = \(\frac { 1 }{ 2 }\) (∠C + ∠D)

Proof: In quadrilateral ABCD

∠A + ∠B + ∠C + ∠D = 360°

\(\frac { 1 }{ 2 }\) (∠A + ∠B + ∠C + ∠D) = 180° …………(ii)

Now in ∆AOB

∠1 + ∠2 + ∠3 = 180° ………(iii)

Equating equation (ii) and equation (iii), we get

∠1 + ∠2 + ∠3 = ∠A + ∠B + \(\frac { 1 }{ 2 }\) (∠C + ∠D)

∠1 + ∠2 + ∠3 = ∠1 + ∠3 + \(\frac { 1 }{ 2 }\) (∠C + ∠D)

∠2 = \(\frac { 1 }{ 2 }\) (∠C + ∠D)

∠AOB = \(\frac { 1 }{ 2 }\) (∠C + ∠D)

Hence proved.