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 cm2
Therefore, the area of a triangle is 5.32 cm2
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 cm2
Hence, the area of a triangle is 486 cm2
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 cm2
Therefore, the area of an equilateral triangle of side 20 cm is 100√3 cm2
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) a2
Given,
A = 16√3
16√3 = (√3 / 4) a2
16 = (a2 / 4)
a2 = 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) a2
A = (√3 / 4) x 62
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 cm2
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 cm2. How long are its sides?
Solution:
Area of a square = 36 cm2
(Side)2 = 36 cm2
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 mm2
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 cm2
Area of a rectangle = 15 x 100 mm2
Area of a rectangle = 1500 mm2
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 cm2 + 9 cm2
= 33 cm2
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 cm2 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) cm2 – [(2 x 4) cm2 + (2)2 cm2]
= 72 cm2 – [8 cm2 + 4 cm2]
= 72 cm2 – 12 cm2
We get,
= 60 cm2
Therefore, the area of the shaded region in the given figure is 60 cm2
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 = 𝝅r2
Hence,
The area of a circle with radius 2.8 cm = 𝝅 (2.8)2
= (22 / 7) (2.8)2
We get,
= 24.64 cm2
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 = 𝝅r2
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 cm2
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 = 𝝅r2
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 cm2
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 = 𝝅r2
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 cm2
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 = (𝝅r2) / 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 cm2
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 = 𝝅r2 / 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 cm2
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 = (𝝅r2 / 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 cm2
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 = 𝝅r2
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 m2
14. Find the circumference of a circle whose area is 81 𝝅 cm2
Solution:
The area of a circle with radius r = 𝝅r2
Here,
Given area of a circle = 81 𝝅 cm2
81 𝝅 = 𝝅r2
r2 = 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 = 𝝅r2 = (22 / 7) x 105 x 105
On calculating, we get,
= 34650 cm2
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