**Frank Solutions for Class 9 Maths Chapter 24 Perimeter and Area** deal with an important concept used in everyday life. It is essential to understand each and every topic thoroughly in Class 9, as this will help to understand the topics in higher classes. In order to come out with flying colours in the annual exam, students are suggested to practise Franks Solutions as many times as possible. It will help them boost their conceptual knowledge as well.

Chapter 24, Perimeter and Area, provides information related to the perimeter and area of a given figure. Perimeter is the distance of the boundary of the shape, and the area is the region occupied by it. Students who refer to the Frank Solutions while revising the textbook problems can master the concepts covered in this chapter undoubtedly. Download Frank Solutions for Class 9 Maths Chapter 24 Perimeter and Area PDF from the links available below.

## Frank Solutions for Class 9 Maths Chapter 24 Perimeter and Area Download PDF

## Access Frank Solutions for Class 9 Maths Chapter 24 Perimeter and Area

**1. Find the area of a triangle whose base is 3.8 cm and height is 2.8 cm.**

**Solution:**

Given

Base of a triangle = 3.8 cm

Height of a triangle = 2.8 cm

We know that,

Area of a triangle = (1/2) x Base x Height

Substituting the values, we get,

= (1/2) x 3.8 x 2.8

= 5.32 cm^{2}

Therefore, the area of a triangle is 5.32 cm^{2}

**2. Find the area of a triangle whose sides are 27 cm, 45 cm and 36 cm.**

**Solution:**

Let the three sides of a triangle be,

a = 27 cm, b = 45 cm and c = 36 cm

Semi-perimeter of a triangle = s = {(a + b + c) / 2}

s = {(27 + 45 + 36) / 2}

We get,

s = 54 cm

Area of a triangle = √s (s – a) (s – b) (s – c)

= √54 (54 – 27) (54 – 45) (54 – 36)

= √54 x 27 x 9 x 18

This can be written as,

= √6 x 9 x 3 x 9 x 9 x 6 x 3

= √6 x 6 x 3 x 3 x 9 x 9 x 9

On calculating further, we get,

= 6 x 3 x 9 x 3

We get,

= 486 cm^{2}

Hence, the area of a triangle is 486 cm^{2}

**3. Find the area of an equilateral triangle of side 20 cm.**

**Solution:**

Given

Side of an equilateral triangle = 20 cm

Area of a triangle = (√3 / 4) x (side)^{2}

= (√3 / 4) x 20 x 20

We get,

= 100√3 cm^{2}

Therefore, the area of an equilateral triangle of side 20 cm is 100√3 cm^{2}

**4. Find the perimeter of an equilateral triangle whose area is 16√3 cm.**

**Solution:**

We know that,

Area of an equilateral triangle of side ‘a’ is,

A = (√3 / 4) a^{2}

Given,

A = 16√3

16√3 = (√3 / 4) a^{2}

16 = (a^{2} / 4)

a^{2} = 4 x 16

We get,

a = 2 x 4

a = 8

Hence, the side of an equilateral triangle is 8 cm

The perimeter of an equilateral triangle of side a = 3a

= 3 x 8

= 24 cm

Therefore, the perimeter of an equilateral triangle of side 8 cm is 24 cm

**5. Find the area of an equilateral triangle having a perimeter of 18 cm.**

**Solution:**

We know that,

The perimeter of an equilateral triangle (P) of side a = 3a

Here,

P = 18 cm

Side of the equilateral triangle = 6 cm

The area of an equilateral triangle (A) of side ‘a’ is A = (√3 / 4) a^{2}

A = (√3 / 4) x 6^{2}

A = (√3 / 4) x 36

We get,

A = 9√3

Hence, the area of an equilateral triangle (A) of side 6 cm is 9√3 cm^{2}

**6. The side of a square is of length 20 mm. Find its perimeter in cm.**

**Solution:**

Given

Side of a square = 20 mm

We know,

Perimeter of square = 4 x side

Perimeter of square = 4 x 20 mm

We get,

Perimeter of square = 80 mm

Perimeter of square = (80 / 10) cm

Perimeter of square = 8 cm

Therefore, the perimeter is 8 cm

**7. The area of a square is 36 cm ^{2}. How long are its sides?**

**Solution:**

Area of a square = 36 cm^{2}

(Side)^{2} = 36 cm^{2}

Side = √36 cm

We get,

Side = 6 cm

Therefore, the length of each side is 6 cm

**8. The sides of a rectangle are 5 cm and 3 cm, respectively. Find its area in mm ^{2}**

**Solution:**

Given

Length of a rectangle = 5 cm

Breadth of a rectangle = 3 cm

We know that,

Area of a rectangle = Length x Breadth

Area of a rectangle = 5 cm x 3 cm

Area of a rectangle = 15 cm^{2}

Area of a rectangle = 15 x 100 mm^{2}

Area of a rectangle = 1500 mm^{2}

**9. Find the area and perimeter of the given figure.**

**Solution:**

Area of given figure = Area of rectangle ABCH + Area of square DEFG

= AB x BC + (DE)^{2}

On substituting, we get,

= 8 cm x 3 cm + (3 cm)^{2}

= 24 cm^{2} + 9 cm^{2}

= 33 cm^{2}

Perimeter of the given figure = AB + BC + CD + DE + EF + FG + GH + HA

Perimeter of the given figure = AB + BC + (CD + EF + GH) + DE + FG + HA

Perimeter of the given figure = 8 + 3 + 8 + 3 + 3 + 3

We get,

The perimeter of the given figure = 28 cm

Therefore, the area and perimeter of the given figure are 33 cm^{2} and 28 cm

**10. Find the shaded area in the given figure.**

**Solution:**

Area of shaded region = Area of rectangle PQRS – (Area of rectangle ABFG + Area of square CDEF)

= PQ x QR – [(AB x AG) + (CD)^{2}]

= (8 x 9) cm^{2} – [(2 x 4) cm^{2} + (2)^{2} cm^{2}]

= 72 cm^{2} – [8 cm^{2} + 4 cm^{2}]

= 72 cm^{2} – 12 cm^{2}

We get,

= 60 cm^{2}

Therefore, the area of the shaded region in the given figure is 60 cm^{2}

**11. Find the area and perimeter of the circles with the following:**

**(i) Radius = 2.8 cm**

**(ii) Radius = 10.5 cm**

**(iii) Diameter = 77 cm**

**(iv) Diameter = 35 cm**

**Solution:**

(i) We know,

The area of a circle with radius r = 𝝅r^{2}

Hence,

The area of a circle with radius 2.8 cm = 𝝅 (2.8)^{2}

= (22 / 7) (2.8)^{2}

We get,

= 24.64 cm^{2}

The circumference of a circle with radius r = 2 𝝅r

The circumference of a circle with radius 2.8 cm = 2 x 𝝅 x 2.8

= 2 x (22 / 7) x 2.8

We get,

= 17.6 cm

(ii) The area of a circle with radius r = 𝝅r^{2}

Therefore,

The area of a circle with a radius 10.5 cm = 𝝅 (10.5)^{2}

= (22 / 7) (10.5)^{2}

We get,

= 346.5 cm^{2}

The circumference of a circle with radius r = 2𝝅r

The circumference of a circle with radius 10.5 cm = 2 x 𝝅 x 10.5

= 2 x (22 / 7) x 10.5

We get,

= 66 cm

(iii) The radius of a circle with diameter d is r = (d / 2)

The area of a circle with radius r = 𝝅r^{2}

The radius of a circle with diameter 77 is r = (77/2)

r = 38.5 cm

The area of a circle with radius r = 𝝅 (38.5)^{2}

r = (22/7) x (38.5)^{2}

r = 4658.5 cm^{2}

The circumference of a circle with diameter d is 𝝅d

The circumference of a circle with diameter 77 is 𝝅 x 77

= (22 / 7) x 77

We get,

= 242 cm

(iv) The radius of a circle with diameter d is r = (d / 2)

The area of a circle with radius r = 𝝅r^{2}

The radius of a circle with diameter 35 is r = (35 / 2) = 17.5 cm

The area of a circle with radius r = 𝝅 (17.5)^{2}

= (22 / 7) x (17.5)^{2}

We get,

= 962.5 cm^{2}

The circumference of a circle with diameter d is 𝝅d

The circumference of a circle with diameter 35 is 𝝅 x 35 = (22 / 7) x 35

= 110 cm

**12. Find the area and perimeter of the following semicircles:**

**(i) Radius = 1.4 cm**

**(ii) Diameter = 7 cm**

**(iii) Diameter = 5.6 cm**

**Solution:**

(i) The area of a semi-circle with radius r = (𝝅r^{2}) / 2

The perimeter of a semi-circle with radius r = 𝝅r + 2r

= r (𝝅 + 2) {By taking ‘r’ as common}

= r {(22/ 7) + 2}

We get,

= (36 / 7) x r

Given radius = 1.4 cm

The area of a semi-circle with radius 1.4 cm = {𝝅 x (1.4)^{2} / 2}

= {(22/ 7) x (1.4)^{2} / 2}

= 3.08 cm^{2}

The perimeter of a semi-circle with radius r = 𝝅r + 2r

= 1.4 (𝝅 + 2)

= 1.4 {(22 / 7) + 2}

= (36 / 7) x 1.4

We get,

= 7.2 cm

(ii) The radius of a circle with diameter d is r = (d / 2)

The area of a semi-circle with radius r = 𝝅r^{2} / 2

The perimeter of a semi-circle with radius r = 𝝅r + 2r

= r (𝝅+ 2)

= r {(22 / 7) + 2}

= (36 / 7) x r

The radius of a circle with diameter 7 is r = (7/ 2)

= 3.5 cm

The area of a semi-circle with radius 3.5 = (3.5)^{2} / 2

= (22/7) x (3.5)^{2} / 2

We get,

= 19.25 cm^{2}

The perimeter of a semi-circle with radius r = 𝝅 x 3.5 + 2 x 3.5

= 3.5 (𝝅 + 2)

= 3.5 {(22/ 7) + 2}

= (36 / 7) x 3.5

We get,

= 18 cm

(iii) The radius of a circle with diameter d is r = (d / 2)

The area of a semi-circle with radius r = (𝝅r^{2} / 2)

The perimeter of a semi-circle with radius r = 𝝅r + 2r

= r (𝝅 + 2)

= r {(22/ 7) + 2}

= (36 / 7) x r

The radius of a circle with diameter 5.6 is r = (5.6) / 2

= 2.8 cm

The area of a semi-circle with radius 2.8 = {(2.8)^{2}} / 2

We get,

= 12.32 cm^{2}

The perimeter of a semi-circle with radius r = 𝝅 x 2.8 + 2 x 2.8

= 2.8 (𝝅 + 2)

= 2.8 {(22/7) + 2}

= (36 / 7) x 2.8

We get,

= 14.4 cm

**13. Find the area of a circular field that has a circumference of 396 m.**

**Solution:**

The circumference of a circle with radius r = 2 𝝅r

Here,

Given, the circumference of a circle = 396 m

2 𝝅r = 396

r = 396 / 2 𝝅

r = (396 x 7) / (2 x 22)

We get,

r = 2772 / 44

r = 63 m

We know,

The area of a circle with radius r = 𝝅r^{2}

Hence,

The area of a circle with radius 63 m = (63)^{2} = (22/ 7) x (63)^{2}

= (22 / 7) x 3969

We get,

= 12, 474 m^{2}

**14. Find the circumference of a circle whose area is 81 𝝅 cm ^{2}**

**Solution:**

The area of a circle with radius r = 𝝅r^{2}

Here,

Given area of a circle = 81 𝝅 cm^{2}

81 𝝅 = 𝝅r^{2}

r^{2} = 81

We get,

r = 9 cm

The circumference of a circle with radius r = 2 𝝅r

The circumference of a circle with radius 9 = 2 𝝅 x 9

We get,

= 18 𝝅 cm

**15. The circumference of a circle exceeds its diameter by 450 cm. Find the area of the circle.**

**Solution:**

Let the radius of a circle = r cm

Circumference of a circle = 2 𝝅r cm

Diameter of a circle = 2r cm

Given,

Circumference of a circle – Diameter of a circle = 450 cm

2 𝝅r – 2r = 450

Taking 2r common, we get,

2r (𝝅 – 1) = 450

2r {(22/7) – 1} = 450

2r (15 / 7) = 450

r = (450 x 7) / (2 x 15)

r = 3150 / 30

We get,

r = 105 cm

Hence,

Area of a circle = 𝝅r^{2} = (22 / 7) x 105 x 105

On calculating, we get,

= 34650 cm^{2}

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