Concise Selina Solutions for Class 9 Maths Chapter 20 - Area and Perimeter of Plane Figures

Selina Solutions for Class 9 Maths Chapter 20 Area and Perimeter of Plane Figures are provided here. It is crucial to understand the concepts taught in Class 9 as these concepts pick up the threads in Class 10. It is advised to solve questions provided in each exercise across all the chapters in the book by Selina publication to score exceptionally well in Class 9 Mathematics examination. This Selina solutions for Class 9 Maths helps students in mastering all the concepts in an effective way. Download pdf of Class 9 Maths Chapter 20 Selina Solutions from the link given below.

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Exercise 20(A)

1. Find the area of a triangle whose sides are 18 cm, 24 cm and 30 cm.

Also, find the length of altitude corresponding to the largest side of the triangle.

Solution:

The sides of the triangle are 18 cm, 24 cm and 30 cm respectively

The semi-perimeter is

s = (18 + 24 + 30)/2

= 36

Hence, the area of the triangle is

A = √[s(s – a)(s – b)(s – c)]

= √[36(36 – a)(36 – 24)(36 – 30)]

= √(36 x 18 x 12 x 6)

= √46656

= 216 sq. cm

Again, we have

Area = ½ x base x altitude

Hence,

216 = ½ x 30 x h

h = 14.4 cm

2. The length of the sides of a triangle are in the ratio 3: 4: 5. Find the area of the triangle if its perimeter is 144 cm.

Solution:

Let’s assume the sides of the triangle to be

a = 3x, b = 4x and c = 5x

And, given that the perimeter is 144 cm

So,

3x + 4x + 5x = 144

12x = 144

x = 144/12

x = 12 … (i)

Now, semi-perimeter is

s = (a + b + c)/2

= 12x/2

= 6x

= 6(12) … [From (i)]

= 72

So, the sides of the triangle are a = 36 cm, b = 48 cm and c = 60 cm

Hence, the area of the triangle is

A = √[s(s – a)(s – b)(s – c)]

= √[72(72 – 36)(72 – 48)(72 – 60)]

= √(72 x 36 x 24 x 12)

= √746496

= 864 cm²

3. ABC is a triangle in which AB = AC = 4 cm and ∠A = 90^o. Calculate:

(i) The area of ∆ABC,

(ii) The length of perpendicular from A to BC.

Solution:

(i) Area of the triangle is given by

A = ½ x AB x AC

= ½ x 4 x 4

= 8 sq. cm

(ii) Now, again the area of the triangle can be expressed as

A = ½ x BC x h

So,

8 = ½ x √(4² + 4²) x h [By Pythagoras theorem, BC² = 4² + 4² ]

8 = ½ x 4√2 x h

8 = 2√2 x h

h = 8/ 2√2

h = 2.83 cm

Hence, the length of perpendicular from A to BC is 2.83 cm

4. The area of an equilateral triangle is 36√3 sq. cm. Find its perimeter.

Solution:

Given, area of equilateral triangle = 36√3 cm²

We know that,

Area of an equilateral triangle is given by,

A = √3/4 x (side)²

So, now

√3/4 x (side)² = 36√3 

(side)² = (36√3)/(√3/4)

= 36 x 4

= 144

Taking square root on both sides, we get

Side of the equilateral triangle = 12 cm

Hence, the perimeter of the equilateral triangle = 3 x side

= 3 x 12

= 36 cm

5. Find the area of an isosceles triangle whose perimeter is 36 cm and base is 16 cm.

Solution:

Given,

Perimeter of the isosceles triangle = 36cm and base = 16cm

Since, the length of two sides are equal

The sides are (36 – 16)/2 = 10 cm each 

Now, we have

a = equal sides = 10 cm and

b = base = 16 cm

Let’s assume ‘h’ to be the altitude of the isosceles triangle.

As the altitude from the vertex bisects the base perpendicularly, we can apply Pythagoras Theorem.

Hence, we have

h = √[a² – (b/2)²]

= ½ √(4a² – b²)

And, the area of the triangle is given by

A = ½ x base x altitude

= ½ x b x ½ √(4a² – b²)

= ½ x 16 x ½ x √(4 x 10² – 16²)

= ¼ x 16 x √(400 – 256)

= 4 x √144

= 4 x 12

= 48 sq. cm

Therefore, the area of the isosceles triangle is 48 sq. cm

Exercise 20(B)

1. Find the area of a quadrilateral one of whose diagonals is 30 cm long and the perpendiculars from the other two vertices are 19 cm and 11 cm respectively.

Solution:

We know that,

Area of quadrilateral = ½ x one diagonal x (sum of the lengths of the perpendiculars drawn from it on the remaining two vertices)

= ½ x 30 x (11 + 19)

= 15 x 30

= 450 sq. cm

2. The diagonals of a quadrilateral are 16 cm and 13 cm. If they intersect each other at right angles; find the area of the quadrilateral.

Solution:

We know that,

Area of the quadrilateral = ½ x the product of the diagonals

= ½ x 16 x 13

= 8 x 13

= 104 cm²

3. Calculate the area of quadrilateral ABCD, in which ∠ABD = 90^o, triangle BCD is an equilateral triangle of side 24 cm and AD = 26 cm.

Solution:

Let’s consider the below figure:

From the right triangle ABD, we have ∠ABD = 90^o

So, by Pythagoras Theorem

AB = √(26² – 24²)

= √(676 – 576)

= √100

= 10 cm

Now, the area of right triangle ABD is

Ar (∆ABD) = ½ x AB x BD

= ½ x 10 x 24

= 120 cm²

Again, in the equilateral triangle BCD we have, CP ⊥ BD

So, by Pythagoras Theorem

PC = √(24² – 12²)

= √(576 – 144)

= √432

= √(144 x 3)

= 12√3 cm

Now, the area of the triangle BCD is

Ar (∆BCD) = ½ x BD x PC

= ½ x 24 x 12√3

= 144√3 cm²

Therefore, the area of the quadrilateral is given by

Ar(ABCD) = Ar (∆ABD) + Ar (∆BCD)

= (120 + 144√3) cm²

= 369.41 cm²

4. Calculate the area of quadrilateral ABCD in which AB = 32 cm, AD = 24 cm, ∠A = 90^o and BC = CD = 52 cm.

Solution:

The figure can be drawn as follows:

We have, quadrilateral ABCD in which AB = 32 cm, AD = 24 cm, ∠A = 90^o and BC = CD = 52 cm

Now, as ABD is a right triangle, it’s the area is given as

∆ABD = ½ x 24 x 32

= 12 x 32

= 384 cm²

Again, by Pythagoras Theorem

BD = √(24² + 32²)

= 8√(3² + 4²)

= 8√25

= 8 x 5

= 40 cm

Now, as BCD is an isosceles triangle and BP ⊥ BD, we have

DP = ½ BD

= ½ x 40

= 20 cm

Then,

From the right triangle DPC, we have

PC = √(52² – 20²) [By Pythagoras Theorem]

= 4√(13² – 5²)

= 4√(169 – 25)

= 4 x √144

= 4 x 12

= 48 cm

So, the area of ∆DPC = ½ x 40 x 48

= 20 x 48

= 960 cm²

Therefore, the area of the quadrilateral is given by

Ar (∆ABD) + Ar (∆DPC) = 960 + 384

= 1344 cm²

5. The perimeter of a rectangular field is 3/5 km. If the length of the field is twice its width; find the area of the rectangle in sq. metres.

Solution:

Let’s assume the width of the rectangular field to be x km and length to be 2x km

Now, according to the question, we have

2(x + 2x) = 3/5

3x = 3/10

x = 1/10 km

i.e., x = 1000/10 m = 100 m

Thus, the width is 100 m and the length is 200m of the rectangular field

Therefore, the area of the rectangular field is

A = length x width

= 100 x 200

= 20,000 sq. m

Exercise 20(C)

1. The diameter of a circle is 28 cm. Find its:

(i) Circumference

(ii) Area.

Solution:

Let’s assume r to be the radius of the circle

(i) Given, diameter = 28 cm

So, radius = 28/2 = 14 cm

Now,

Circumference = 2πr

= 2 x 22/7 x 14

= 88 cm

(ii) The area of the circle is given by

Area = πr²

= 22/7 x 14²

= 22/7 x 14 x 14

= 44 x 14

= 616 cm²

2. The circumference of a circular field is 308 m. Find is:

(i) Radius

(ii) Area.

Solution:

Let’s assume r to the radius of the circular field

(i) Given,

The circumference of the circular field = 308 m

2πr = 308

r = 308/2π

= 154/π

= (154 x 7)/22

= 49 m

Hence, the radius of the circular field is 49 m

(ii) Now, the area of the circular field is calculated as

Area = πr²

= 22/7 x 49²

= 22/7 x 49 x 49

= 22 x 7 x 49

= 7546 cm²

3. The sum of the circumference and diameter of a circle is 116 cm. Find its radius.

Solution:

Let’s consider r to be the radius of the circle

Then, according to the question, we have

2πr + 2r = 116

2r(π + 1) = 116

r = 116/2(π + 1)

= 88/(22/7 + 1)

= 14 cm

Hence, the radius of the circle is 14 cm

4. The radii of two circles are 25 cm and 18 cm. Find the radius of the circle which has circumference equal to the sum of circumferences of these two circles.

Solution:

We have,

The radii of two circles are 25 cm and 18 cm

Now, the circumference of the first circle is

S₁ = 2π x 25

= 50π cm

And,

The circumference of the second circle is

S_s = 2π x 18

= 36π cm

According to the question,

Let’s assume R to be the radius of the resulting circle

So,

2πR = 50π + 36π

2πR = π(50 + 36)

Dividing by π on both sides, we get

2R = 86

R = 43 cm

Therefore, the radius of the circle which has circumference equal to the sum of circumferences of the given two circles is 43 cm

5. The radii of two circles are 48 cm and 13 cm. Find the area of the circle which has its circumference equal to the difference of the circumferences of the given two circles.

Solution:

We have,

The radii of two circles are 48 cm and 13 cm

Now, the circumference of the first circle is

S₁ = 2π x 48

= 96π cm

And,

The circumference of the second circle is

S_s = 2π x 13

= 26π cm

According to the question,

Let’s assume R to be the radius of the resulting circle

So,

2πR = 96π – 26π

2πR = π(96 – 26)

Dividing by π on both sides, we get

2R = 70

R = 35 cm

Then, the area of the circle is given by

A = πR²

= π x 35²

= 22/7 x 35 x 35

= 22 x 5 x 35

= 3850 cm²

Therefore, the area of the required circle is 3850 cm²

Exercise 20(D)

1. The perimeter of a triangle is 450 cm and its side are in the ratio 12 : 5 : 13. Find the area of the triangle.

Solution:

Let’s assume the sides of the triangle to be

a = 12x

b = 5x

c = 13x

And, given that the perimeter of the triangle = 450 cm

So, 12x + 5x + 13x = 450

⇒ 30x = 450

⇒ x = 15

Thus, the sides of a triangle are

a = 12x = 12(15) = 180 cm

b = 5x = 5(15) = 75 cm

c = 13x = 13(15) = 195 cm

Now,

The semi-perimeter of the triangle, s = (a + b + c)/2

= (180 + 75 + 195)/2

= 450/2

= 225 cm

Hence, the area of the triangle is given by

Area = √[s(s – a)(s – b)(s – c)]

= √[225(225 – 180)(225 – 75)(225 – 195)]

= √(225 x 45 x 150 x 30)

= 15 √(9 x 5 x 5 x 30 x 30)

= 15 x 3 x 5 x 30

= 6750 cm²

2. A triangle and a parallelogram have the same base and the same area. If the side of the triangle are 26 cm, 28 cm and 30 cm and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Solution:

Let’s assume the sides of the triangle to be

a = 26 cm, b = 28 cm and c = 30 cm

Now,

The semi-perimeter of the triangle, s = (a + b + c)/2

= (26 + 28 + 30)/2

= 84/2

= 42 cm

Hence, the area of the triangle is given by

Area = √[s(s – a)(s – b)(s – c)]

= √[42(42 – 26)(42 – 28)(42 – 30)]

= √(42 x 16 x 14 x 12)

= √(7 x 6 x 4² x 7 x 2 x 6 x 2)

= 7 x 6 x 4 x 2

= 336 cm²

Given, the base of the parallelogram = 28 cm

And, area of the parallelogram = area of the triangle

So,

Base x Height = 336

28 x Height = 336

Height = 336/28

= 12 cm

Therefore, the height of the parallelogram is 12 cm

3. Using the information in the following figure, find its area.

Solution:

Let’s make a construction of drawing CM ⊥ AB

Now, 

In right-angled triangle CMB, we have

BM² = BC² – CM²  [By Pythagoras Theorem]

= (15)² – (9)² 

= 225 – 81

= 144 m

On taking square root on both sides, we get

BM = 12 m

Now,

AB = AM + BM

= 23 + 12

= 35 m

Hence, the area of the trapezium = ½ x (sum of parallel sides) x Height

= ½ x (AB + CD) x AD

= ½ x (23 + 35) x 9

= ½ x 58 x 9

= 261 m²

4. Sum of the areas or two squares is 400 cm². If the difference of their perimeters is 16 cm, find the sides of the two squares.

Solution:

Let’s assume the sides of the two squares to be a and b respectively

Then,

Area of one square, S₁ = a²

And, area of the second square, S₂ = b²

According to the question, we have

S₁ + S₂ = 400 cm²

⇒ a² + b² = 400 cm² … (1)

Also given, the difference in their perimeters = 16 cm

⇒ 4a – 4b = 16 cm

a – b = 4

a = (4 + b)

Substituting the value of ‘a’ in (1), we get

(4 + b)² + b² = 400

16 + 8b + b² + b² = 400

2b² + 8b – 384 = 0

b² + 4b – 192 = 0

b² + 16b – 12b – 192 = 0

b(b + 16) – 12(b + 16) = 0

(b +16) (b – 12) = 0

b + 16 = 0 or b – 12 = 0

⇒ b = -16 or b = 12

As, the side of a square cannot be negative, we neglect the value -16.

Hence, b = 12

And,

a = 4 + b = 4 + 12 = 16

Therefore, the sides of a square are 16 cm and 12 cm respectively

5. Find the area and the perimeter of a square with diagonal 24 cm. [Take √2 = 1.41]

Solution:

Given, the diagonal of a square = 24 cm

We know that,

Diagonal of a square = √2 times the side of a square

24 = √2 x (side of a square)

So,

Side of the square = 24/√2

= 12√2 cm

Thus,

The perimeter of the square = 4 x side

= 4 x 12√2

= 48√2

= 48 x 1.41

= 67.68 cm

And,

The area of the square = (side)²

= (12√2)²

= 144 x 2

= 288 cm²

Selina Solutions for Class 9 Maths Chapter 20 – Area and Perimeter of Plane Figures

The Chapter 20, Area and Perimeter of Plane Figures, has 4 exercises and the Selina Solutions given here contains the answers to all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.

20.1 Introduction

20.2 Area and perimeter of triangles.

20.3 Some special types of triangles

20.4 Area and Perimeter of quadrilaterals

20.5 Some special types of quadrilaterals

20.6 Circumference of a circle

20.7 Area of a circle

Selina Solutions for Class 9 Maths Chapter 20-Area and Perimeter of Plane Figures

The perimeter of a plane figure is the length of its boundary. On the other hand, the area of a plane figure is the measure of the surface enclosed by its boundary. Chapter 20 of class 9 gives the students an overview of the Area as well as the Perimeter of Plane Figures. Read and learn Chapter 20 of Selina textbook to learn more about Area and Perimeter of Plane Figures, along with the concepts covered in it. Solve the Selina Solutions for Class 9 effectively to score high in the examination.