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 cm^{2} = 24.48 cm^{2}

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 cm^{2} = 38.64 cm^{2}

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 cm^{2} = 17.5 cm^{2}

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 cm^{2} 4 × 4 cm^{2} = 16 cm^{2}

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 cm^{2}

= 3.1 × 4.3 cm^{2} = 13.33 cm^{2}

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 cm^{2} = 4 × 6 cm^{2} = 24 cm^{2}

Question 7.

Solution:

In the circle, diameter, d = 14 cm

radius, r = \(\frac { d }{ 2 }\) = \(\frac { 14 }{ 2 }\) = 7 cm

Area of circle = πr^{2}

= \(\frac { 22 }{ 7 }\) × 7 × 7 cm^{2}

= 22 × 7 cm^{2} = 154 cm^{2}

Question 8.

Solution:

In the circle, radius, r = 3.5 cm

Area of circle = πr^{2}

= \(\frac { 22 }{ 7 }\) × 3.5 × 3.5 cm^{2}

= 22 × 0.5 × 3.5 cm^{2}

= 11 × 3.5 cm^{2} = 38.5 cm^{2}

Question 9.

Solution:

In a semicircle diameter, d = 21 cm

radius, r = \(\frac { d }{ 2 }\) = \(\frac { 21 }{ 2 }\) = cm

Area of semicircle = \(\frac{\pi r^2}{2}\)

= \(\frac { 22 }{ 7 }\) × \(\frac { 1 }{ 2 }\) × \(\frac { 21 }{ 2 }\) × \(\frac { 21 }{ 2 }\) cm^{2}

= 11 × \(\frac { 3 }{ 2 }\) × \(\frac { 21 }{ 2 }\) cm^{2} = 173.25 cm^{2}

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 cm^{2} = 3 × 3 cm^{2} = 9 cm^{2}

Area of parallelogram = 2 × area of triangle

=2 × 9 cm^{2} = 18 cm^{2}

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 = πr^{2}

= \(\frac { 22 }{ 7 }\) × 1.4 × 1.4 cm^{2}

= 22 × 0.2 × 1.4 cm^{2} = 6.16 cm^{2}

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 x^{2} = 8^{2} + 6^{2}

x^{2} = 64 + 36

x^{2} = 100 ⇒ x^{2} = 10^{2} = 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 m^{2}

in a rectangle

l = 240m, b = 120m

Area of rectangle l × b

= 240 × 120 m^{2} = 28,800 m^{2}

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 × 110m^{2} = 12,100 m^{2}

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 cm^{2} = 800 cm^{2}

Question 17.

Two diagonals of a Rhombus are 8 cm and 6 cm. Find its area.

Solution:

We know that area of Rhombus A = \(\frac { 1 }{ 2 }\)d_{1}d_{2}

Given d_{1} = 8 cm : d_{2} = 6 cm

A = \(\frac { 1 }{ 2 }\) × 8 × 6 cm^{2}

A = 4 × 6 cm^{2} = 24 cm^{2}

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 cm^{2} = 19.2 cm^{2}

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 cm^{2} = 25.2 cm^{2}

Question 20.

Find the height of a triangle whose base is 20 cm and area Is 180 cm^{2}.

Solution:

Area of triangle, A = \(\frac { 1 }{ 2 }\) × b × h

Given base, b = 20 cm: A = 180 cm^{2}

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 m^{2}. Find its perimeter.

Solution:

Given, area of square = 1024 m^{2}

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 cm^{2}, then find z.

Solution:

Given area of parallelogram = 144 cm^{2}

Given base b = 9 cm

height = h cm

∴ b × h = 144 cm^{2}

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 cm^{2}. Find its diameter.

Solution:

Given area of circle = 616 cm^{2}

πr^{2} = 616

\(\frac { 22 }{ 7 }\) × r = 616

r^{2} = 616 × \(\frac { 7 }{ 22 }\)

r^{2} = 196 = 14^{2} ⇒ 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 = πR^{2}

= π(3r)^{2} = π9r^{2} = 9πr^{2} 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 = πr^{2}

= \(\frac { 22 }{ 7 }\) × 14 × 14 cm^{2} = 22 × 2 × 14 cm^{2}

= 44 × 14 cm^{2} = 616 cm^{2}

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 cm^{2}. Find the length of other diagonal.

Solution:

In a Rhombus, Diagonal d_{1} = 16 cm

Area = 96 cm^{2}

\(\frac { 1 }{ 2 }\) × d_{1} × d_{2} = 96 cm^{2}

\(\frac { 1 }{ 2 }\) × 16 × d_{2} = 96 cm^{2}

8 × d_{2} = 96 cm^{2}

d_{2} = \(\frac { 96 }{ 8 }\) = 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 cm^{2}

= 142800 cm^{2} = 1428 m^{2}

Given cost of painting = ₹10 per m^{2}

Total cost = ₹ 10 × 14.28

= ₹ 142.8 = ₹143.

Question 32.

The area of a parallelogram is 363 cm^{2}. 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 cm^{2}

b × h = 363 cm^{2}

x × 3x = 363 cm^{2}

3x^{2} = 363

x^{2} = \(\frac { 363 }{ 3 }\) ⇒ x^{2} = 121 = 11^{2}

∴ 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 cm^{2}

= 10 × 20 cm^{2} = 200 cm^{2}

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 cm^{2} = 100 cm^{2}

Smaller square

Side of square = 3 cm

Area 3 × 3 cm^{2} = 9 cm^{2}

Area of shaded region = Area of bigger

square — Area of smaller square = 100 – 9 = 91 cm^{2}

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

πR^{2} = 100 πr^{2}

R^{2} = 100 r^{2}

R^{2} = (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