These Class 7 Maths Extra Questions Chapter 9 Perimeter and Area will help students prepare well for the exams.
Class 7 Maths Chapter 9 Extra Questions Perimeter and Area
Class 7 Maths Perimeter and Area Extra Questions – 2 Marks
Question 1.
Solution:
In the parallelogram base b = 7.2 cm
height, h = 3.4 cm
Area of parallelogram = b × h
= 7.2 × 3.4 cm2 = 24.48 cm2
Question 2.
Solution:
In the parallelogram base b = 9.2 cm
height, h = 4.2 cm
Area of parallelogram = b × h
= 9.2 × 4.2 cm2 = 38.64 cm2
Question 3.
Solution:
In the triangle base, b = 7 cm
height, h = 5 cm
Area of triangle = \(\frac { 1 }{ 2 }\) × b × h
= \(\frac { 1 }{ 2 }\) × 7 × 5 cm2 = 17.5 cm2
Question 4.
Solution:
In the triangle base, b = 8 cm
height, h = 4 cm
Area of triangle = \(\frac { 1 }{ 2 }\) × b × h
= \(\frac { 1 }{ 2 }\) × 8 × 4 cm2 4 × 4 cm2 = 16 cm2
Question 5.
Solution:
In the triangle base, b = 6.2 cm
height, h = 4.3 cm
Area of triangle = \(\frac { 1 }{ 2 }\) × b × h
= \(\frac { 1 }{ 2 }\) × 6.2 × 4.3 cm2
= 3.1 × 4.3 cm2 = 13.33 cm2
Question 6.
Solution:
In the triangle base, b = 8 cm
height, h = 6 cm
Area of triangle = \(\frac { 1 }{ 2 }\) × b × h
= \(\frac { 1 }{ 2 }\) × 8 × 6 cm2 = 4 × 6 cm2 = 24 cm2
Question 7.
Solution:
In the circle, diameter, d = 14 cm
radius, r = \(\frac { d }{ 2 }\) = \(\frac { 14 }{ 2 }\) = 7 cm
Area of circle = πr2
= \(\frac { 22 }{ 7 }\) × 7 × 7 cm2
= 22 × 7 cm2 = 154 cm2
Question 8.
Solution:
In the circle, radius, r = 3.5 cm
Area of circle = πr2
= \(\frac { 22 }{ 7 }\) × 3.5 × 3.5 cm2
= 22 × 0.5 × 3.5 cm2
= 11 × 3.5 cm2 = 38.5 cm2
Question 9.
Solution:
In a semicircle diameter, d = 21 cm
radius, r = \(\frac { d }{ 2 }\) = \(\frac { 21 }{ 2 }\) = cm
Area of semicircle = \(\frac{\pi r^2}{2}\)
= \(\frac { 22 }{ 7 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 21 }{ 2 }\) × \(\frac { 21 }{ 2 }\) cm2
= 11 × \(\frac { 3 }{ 2 }\) × \(\frac { 21 }{ 2 }\) cm2 = 173.25 cm2
Question 10.
Solution:
In the triangle base, b = 6 cm
height, h = 3 cm
Area of triangle = \(\frac { 1 }{ 2 }\) × b × h
= \(\frac { 1 }{ 2 }\) × 6 × 3 cm2 = 3 × 3 cm2 = 9 cm2
Area of parallelogram = 2 × area of triangle
=2 × 9 cm2 = 18 cm2
Question 11.
Find the area and circumference of a circle of radius 1.4 cm. (Take π = \(\frac { 22 }{ 7 }\))
Solution:
In a circle radius, r = 1.4cm
Area of circle = πr2
= \(\frac { 22 }{ 7 }\) × 1.4 × 1.4 cm2
= 22 × 0.2 × 1.4 cm2 = 6.16 cm2
Circumference = 2πr
= 2 × \(\frac { 22 }{ 7 }\) × 1.4 cm
= 44 × 0.2 cm = 8.8cm
Question 12.
A rectangle has the measurements of length and breadth as 8 cm and 6 cm. Find its diagonal measure.
Solution:
ABCD is a rectangle
In right triangle ABC x2 = 82 + 62
x2 = 64 + 36
x2 = 100 ⇒ x2 = 102 = 10 cm
∴ Length of diagonal of rectangle = 10 cm
Question 13.
Find the area in hectare of a field whose length is 240 m and breadth is 120 m.
Solution:
I hectare 10,000 sq.m = 10,000 m2
in a rectangle
l = 240m, b = 120m
Area of rectangle l × b
= 240 × 120 m2 = 28,800 m2
Area of rectangle in hectares = \(\frac { 28,800 }{ 10,000 }\)
= \(\frac { 288 }{ 100 }\) = 2.88 hectares.
Question 14.
Find the area of a square park whose perimeter is 440 m.
Solution:
Given perimeter of square park 440 m
4a = 440 ⇒ a = \(\frac { 440 }{ 4 }\) = 110 m
Side of square park = 110m
Area of square park = a × a
= 110 × 110m2 = 12,100 m2
Question 15.
Find the perimeter of below semicircle.
Solution:
In a semicircle
diameter, d = 14 cm
radius, r = \(\frac { d }{ 2 }\) = \(\frac { 14 }{ 2 }\)cm = 7 cm
Perimeter of semicircle = πr + 2r
= \(\frac { 22 }{ 7 }\) × 7 + 2 × 7 = 22 + 14 = 36 cm
Perimeter and Area Extra Questions Class 7 – 3 Marks
Question 16.
The perimeter of a rectangular sheet is 120 cm. If the length Is 40 cm, find its breadth.
Solution:
Given perimeter of rectangle = 120 cm
2(l + b) = 120 cm
l + b = \(\frac { 120 }{ 2 }\) cm
Given length l = 40 cm
1 + b = 60 cm
40 + b= 60 cm
b =60 – 40 = 20 cm
Breadth = 20 cm
Area of rectangle = l × b
= 40 × 20 cm2 = 800 cm2
Question 17.
Two diagonals of a Rhombus are 8 cm and 6 cm. Find its area.
Solution:
We know that area of Rhombus A = \(\frac { 1 }{ 2 }\)d1d2
Given d1 = 8 cm : d2 = 6 cm
A = \(\frac { 1 }{ 2 }\) × 8 × 6 cm2
A = 4 × 6 cm2 = 24 cm2
Question 18.
Find the area of parallelogram with base 6 cm and altitude 3.2 cm.
Solution:
Given in a parallelogram base, b = 6 cm
altitude, h = 3.2 cm
Area of parallejogram = b × h
= 6 × 3.2 cm2 = 19.2 cm2
Question 19.
The side of a Rhombus is 6.3 cm and its altitude Is 4 cm. Find its area.
Solution:
Area of Rhombus = base × altitude = 6.3 × 4 cm2 = 25.2 cm2
Question 20.
Find the height of a triangle whose base is 20 cm and area Is 180 cm2.
Solution:
Area of triangle, A = \(\frac { 1 }{ 2 }\) × b × h
Given base, b = 20 cm: A = 180 cm2
180 = \(\frac { 1 }{ 2 }\) × 20 × h
180 = 10 × h
h = \(\frac { 180 }{ 10 }\) = 18 cm
∴ Height of the triangle = 18 cm
Question 21.
The area of square is 1024 m2. Find its perimeter.
Solution:
Given, area of square = 1024 m2
a × a = 1024
a × a = 32 × 32 ⇒ a = 32 m
Side of square = 32 m
Perimeter of square = 4 × side =4 × 32= 128 m.
Question 22.
In the figure PQRS is a parallelogram and its area is 144 cm2, then find z.
Solution:
Given area of parallelogram = 144 cm2
Given base b = 9 cm
height = h cm
∴ b × h = 144 cm2
9 × h = 144
h = \(\frac { 144 }{ 9 }\) = 16 cm
∴ Height of the parallelogram = 16 cm
Question 23.
The diameter of the car wheel is 77cm. How many revolution will it make to cover 121 km distance?
Solution:
Given diameter of the car wheel = 77cm
d = 77 cm; r = \(\frac { 77 }{ 2 }\)cm.
Circumference of car wheel = 2πr
= 2 × \(\frac { 22 }{ 7 }\) × \(\frac { 77 }{ 2 }\) = 22 × 11 = 242 cm
Total distance covered = 121 km
= 121 × 100000 cm = 12100000 cm
Number of revolution = \(\frac { 12100000 }{ 242 }\) = 50000
Question 24.
Find the diameter of a circle whose circumference is 12.56 cm
Solution:
Given circumference of a circle = 12.56 cm
2πr = 12.56 cm
2 × \(\frac { 22 }{ 7 }\) × r = 12.56
r = 12.56 × \(\frac { 7 }{ 22 }\) × \(\frac { 1 }{ 2 }\)
r = \(\frac { 87.92 }{ 44 }\) = 1.99
Diameter = 2 × r = 2 × 1.99 = 3.98 cm
Question 25.
The area of a circle Is 616 cm2. Find its diameter.
Solution:
Given area of circle = 616 cm2
πr2 = 616
\(\frac { 22 }{ 7 }\) × r = 616
r2 = 616 × \(\frac { 7 }{ 22 }\)
r2 = 196 = 142 ⇒ r = 14cm
Diameter = 2 × r = 2 × 14cm = 28 cm
Extra Questions of Perimeter and Area Class 7 – 5 Marks
Question 26.
Two circles have the ratio of their cir-cuniference as 3 : 2. Find the ratio of their areas.
Solution:
Question 27.
The perimeter of a circle Is 6πr cm. Find its area.
Solution:
Given perimeter of a circle = 6πr cm
2πR = 6πr
2R = 6r ⇒ R = \(\frac { 6r }{ 2 }\) ⇒ R = 3r
Area of circle = πR2
= π(3r)2 = π9r2 = 9πr2 cm.
Question 28.
The difference between the circumference and radius of a circle Is 74 cm. Then find the area of circle.
Solution:
Let r be the radius of circle
According to the problem
2πr – r = 74 cm ⇒ r(2π – 1) = 74
\(r\left(2 \times \frac{22}{7}-1\right)\) = 74 ⇒ \(\left(\frac{44}{7}-1\right)\) = 74
\(r\left(\frac{44-7}{7}\right)\) = 74 ⇒ \(r\left(\frac{37}{7}\right)\) = 74
r = 74 × \(\frac { 7 }{ 37 }\) = 2 × 7 = 14 cm
Area of the circle = πr2
= \(\frac { 22 }{ 7 }\) × 14 × 14 cm2 = 22 × 2 × 14 cm2
= 44 × 14 cm2 = 616 cm2
Question 29.
The length of a rectangular field is thrice Its breadth and Its perimeter is 240 m. Find the length of rectangle.
Solution:
Given in a rectangle
Length is thrice the breadth
Let breadth = b = x m
l length = 3x m
Given perimeter = 240 m
2(l + b) = \(\frac { 240 }{ 2 }\) = 120 m
3x + x = 120m
4x = 120 m ⇒ x = \(\frac { 120 }{ 4 }\) = 30 m
∴ Breadth = 30 m
Length = 3 × x = 3 × 30 = 90 m
Question 30.
The length of one diagonal of a Rhombus is 16 cm. Its area is 96 cm2. Find the length of other diagonal.
Solution:
In a Rhombus, Diagonal d1 = 16 cm
Area = 96 cm2
\(\frac { 1 }{ 2 }\) × d1 × d2 = 96 cm2
\(\frac { 1 }{ 2 }\) × 16 × d2 = 96 cm2
8 × d2 = 96 cm2
d2 = \(\frac { 96 }{ 8 }\) = 12 cm
∴ Length of other diagonal = 12 cm.
Question 31.
A green hoard of sides 4m 20 cm and 3m 40cm is to be painted. Find the cost at the rate of ₹10 per square metre.
Solution:
The shape of green board is rectangle
length l = 4m 20 cm
= (400 + 20) cm = 420 cm
breadth b = 3 m 40 cm
= (300 + 40) cm = 340 cm
Area = l × b = 420 × 340 cm2
= 142800 cm2 = 1428 m2
Given cost of painting = ₹10 per m2
Total cost = ₹ 10 × 14.28
= ₹ 142.8 = ₹143.
Question 32.
The area of a parallelogram is 363 cm2. If its altitude is twice the corresponding base. determine the base and the altitude.
Solution:
Let the base = x cm : Altitude = 3x cm
Area = 363 cm2
b × h = 363 cm2
x × 3x = 363 cm2
3x2 = 363
x2 = \(\frac { 363 }{ 3 }\) ⇒ x2 = 121 = 112
∴ x = 11 cm
Base = 11 cm
Altitude = 3 × 11 = 33 cm
Question 33.
Find the area of an Isosceles Right triangle, if one of the equal sides is 20 cm long.
Solution:
In an Isosceles triangle equal side,
base = 20 cm
Area = \(\frac { 1 }{ 2 }\) × 20 × 20 cm2
= 10 × 20 cm2 = 200 cm2
Question 34.
Find the area of shaded region in the following figure.
Solution:
The diagram consists of two square namely bigger and smaller.
Bigger square
Side of square = 10 cm
Area = 10 × 10 cm2 = 100 cm2
Smaller square
Side of square = 3 cm
Area 3 × 3 cm2 = 9 cm2
Area of shaded region = Area of bigger
square — Area of smaller square = 100 – 9 = 91 cm2
Question 35.
The area of a circle Is 100 times the area of another circle. What is the ratio of their circumferences?
Solution:
Let R and r be the radii of two circles
According to the problem
πR2 = 100 πr2
R2 = 100 r2
R2 = (10r)2
R = 10r
Ratio of circumferences
\(\frac{2 \pi R}{2 \pi r}\) = \(\frac { R }{ r }\) = \(\frac { 10r }{ r }\) = \(\frac { 10 }{ 1 }\) = 10 : 1