## Perimeter and Area Class 7 Extra Questions Maths Chapter 11

**Extra Questions for Class 7 Maths Chapter 11 Perimeter and Area**

### Perimeter and Area Class 7 Extra Questions Very Short Answer Type

**Perimeter And Area Class 7 Worksheets With Answers Question 1.**

The side of a square is 2.5 cm. Find its perimeter and area.

Solution:

Side of the square = 2.5 cm

Perimeter = 4 Ã— Side = 4 Ã— 2.5 = 10 cm

Area = (side)^{2} = (4)^{2} = 16 cm^{2}

**Perimeter And Area Class 7 Extra Questions Question 2.**

If the perimeter of a square is 24 cm. Find its area.

Solution:

Perimeter of the square = 24 cm

Side of the square = \(\frac { 24 }{ 4 }\) cm = 6 cm

Area of the square = (Side)^{2} = (6)^{2} cm^{2} = 36 cm^{2}

**Perimeter And Area Class 7 Questions Question 3.**

If the length and breadth of a rectangle are 36 cm and 24 cm respectively. Find

(i) Perimeter

(ii) Area of the rectangle.

Solution:

Length = 36 cm, Breadth = 24 cm

(i) Perimeter = 2(l + b) = 2(36 + 24) = 2 Ã— 60 = 120 cm

(ii) Area of the rectangle = l Ã— b = 36 cm Ã— 24 cm = 864 cm^{2}

**Class 7 Maths Chapter 11 Extra Questions Question 4.**

The perimeter of a rectangular field is 240 m. If its length is 90 m, find:

(i) it’s breadth

(ii) it’s area.

Solution:

(i) Perimeter of the rectangular field = 240 m

2(l + b) = 240 m

l + b = 120 m

90 m + b = 120 m

b = 120 m – 90 m = 30 m

So, the breadth = 30 m.

(ii) Area of the rectangular field = l Ã— b = 90 m Ã— 30 m = 2700 m^{2}

So, the required area = 2700 m^{2}

**Area And Perimeter Questions For Class 7 Question 5.**

The length and breadth of a rectangular field are equal to 600 m and 400 m respectively. Find the cost of the grass to be planted in it at the rate of â‚¹ 2.50 per m^{2}.

Solution:

Length = 600 m, Breadth = 400 m

Area of the field = l Ã— b = 600 m Ã— 400 m = 240000 m^{2}

Cost of planting the grass = â‚¹ 2.50 Ã— 240000 = â‚¹ 6,00,000

Hence, the required cost = â‚¹ 6,00,000.

**Perimeter And Area Class 7 Worksheet Question 6.**

The perimeter of a circle is 176 cm, find its radius.

Solution:

The perimeter of the circle = 176 cm

**Class 7 Perimeter And Area Extra Questions Question 7.**

The radius of a circle is 3.5 cm, find its circumference and area.

Solution:

Radius = 3.5 cm

Circumference = 2Ï€r

**Perimeter And Area Class 7 Questions With Answers Question 8.**

Area of a circle is 154 cm^{2}, find its circumference.

Solution:

Area of the circle = 154 cm^{2}

**Perimeter And Area Class 7 Worksheets With Answers Pdf Question 9.**

Find the perimeter of the figure given below.

Solution:

Perimeter of the given figure = Circumference of the semicircle + diameter

= Ï€r + 2r

= \(\frac { 22 }{ 7 }\) Ã— 7 + 2 Ã— 7

= 22 + 14

= 36 cm

Hence, the required perimeter = 36 cm.

**Perimeter And Area Questions For Class 7 Question 10.**

The length of the diagonal of a square is 50 cm, find the perimeter of the square.

Solution:

Let each side of the square be x cm.

x^{2} + x^{2} = (50)^{2} [Using Pythagoras Theorem]

2x^{2} = 2500

x^{2} = 1250

x = âˆš1250 = \(\sqrt { 2\times 5\times 5\times 5\times 5 }\)

x = 5 Ã— 5 Ã— âˆš2 = 25âˆš2

The side of the square = 25âˆš2 cm

Perimeter of the square = 4 Ã— side = 4 Ã— 25âˆš2 = 100âˆš2 cm

### Perimeter and Area Class 7 Extra Questions Short Answer Type

**Questions On Area And Perimeter For Class 7 Question 11.**

A wire of length 176 cm is first bent into a square and then into a circle. Which one will have more area?

Solution:

Length of the wire = 176 cm

Side of the square = 176 Ã· 4 cm = 44 cm

Area of the square = (Side)^{2} = (44)^{2} cm^{2} = 1936 cm^{2}

Circumference of the circle = 176 cm

Since 2464 cm^{2} > 1936 cm^{2}

Hence, the circle will have more area.

**Class 7 Perimeter And Area Worksheet Question 12.**

In the given figure, find the area of the shaded portion.

Solution:

Area of the square = (Side)^{2} = 10 cm Ã— 10 cm = 100 cm^{2}

Area of the circle = Ï€r^{2}

= \(\frac { 22 }{ 7 }\) Ã— 3.5 Ã— 3.5

= \(\frac { 77 }{ 2 }\) cm^{2}

= 38.5 cm^{2}

Area of the shaded portion = 100 cm^{2} – 38.5 cm^{2} = 61.5 cm^{2}

**Class 7 Maths Perimeter And Area Extra Questions Question 13.**

Find the area of the shaded portion in the figure given below.

Solution:

Area of the rectangle = l Ã— b = 14 cm Ã— 14 cm = 196 cm^{2}

Radius of the semicircle = \(\frac { 14 }{ 2 }\) = 7 cm

Area of two equal semicircle = 2 Ã— \(\frac { 1 }{ 2 }\) Ï€r^{2}

= Ï€r^{2}

= \(\frac { 22 }{ 7 }\) Ã— 7 Ã— 7

= 154 cm^{2}

Area of the shaded portion = 196 cm^{2} – 154 cm^{2} = 42 cm^{2}

Question 14.

A rectangle park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.

Solution:

Length of the rectangular park = 45 m

Breadth of the park = 30 m

Area of the park = l Ã— 6 = 45m Ã— 30m = 1350 m^{2}

Length of the park including the path = 45 m + 2 Ã— 2.5 m = 50 m

Breadth of the park including the path = 30 m + 2 Ã— 2.5 m = 30m + 5m = 35m

Area of the park including the path = 50 m Ã— 35 m = 1750 m^{2}

Area of the path = 1750 m^{2} – 1350 m^{2} = 400 m^{2}

Hence, the required area = 400 m^{2}.

Question 15.

In the given figure, calculate:

(Ð°) the area of the whole figure.

(b) the total length of the boundary of the field.

Solution:

Area of the rectangular portions = l Ã— b = 80 cm Ã— 42 cm = 3360 cm^{2}

Area of two semicircles = 2 Ã— \(\frac { 1 }{ 2 }\) Ï€r^{2} = Ï€r^{2}

= \(\frac { 22 }{ 7 }\) Ã— 21 Ã— 21

= 22 Ã— 3 Ã— 21

= 1386 cm^{2}

Total area = 3360 cm^{2} + 1386 cm^{2} = 4746 cm^{2}

Total length of the boundary of field = (2 Ã— 80 + Ï€r + Ï€r) cm

= (160 + \(\frac { 22 }{ 7 }\) Ã— 21 + \(\frac { 22 }{ 7 }\) Ã— 21)

= (160 + 132) cm

= 292 cm

Hence, the required (i) area = 4746 cm^{2} and (ii) length of boundary = 292 cm.

Question 16.

How many times a wheel of radius 28 cm must rotate to cover a distance of 352 m?

(Take Ï€ = \(\frac { 22 }{ 7 }\))

Solution:

Radius of the wheel = 28 cm

Circumference = 2Ï€r = 2 Ã— \(\frac { 22 }{ 7 }\) Ã— 28 = 176 cm

Distance to be covered = 352 m or 352 Ã— 100 = 35200 m

Number of rotation made by the wheel to cover the given distance = \(\frac { 35200 }{ 176 }\) = 200

Hence, the required number of rotations = 200.

### Perimeter and Area Class 7 Extra Questions Long Answer Type

Question 17.

A nursery school playground is 160 m long and 80 m wide. In it 80 m Ã— 80 m is kept for swings and in the remaining portion, there are 1.5 m wide path parallel to its width and parallel to its remaining length as shown in Figure. The remaining area is covered by grass. Find the area covered by grass. (NCERT Exemplar)

Solution:

Area of the playground = l Ã— b = 160 m Ã— 80 m = 12800 m^{2}

Area left for swings = l Ã— b = 80m Ã— 80m = 6400 m^{2}

Area of the remaining portion = 12800 m^{2} – 6400 m^{2} = 6400 m^{2}

Area of the vertical road = 80 m Ã— 1.5 m = 120 m^{2}

Area of the horizontal road = 80 m Ã— 1.5 m = 120 m^{2}

Area of the common portion = 1.5 Ã— 1.5 = 2.25 m^{2}

Area of the two roads = 120 m^{2} + 120 m^{2} – 2.25 m^{2} = (240 – 2.25) m^{2} = 237.75 m^{2}

Area of the portion to be planted by grass = 6400 m^{2} – 237.75 m^{2} = 6162.25 m^{2}

Hence, the required area = 6162.25 m^{2}.

Question 18.

Rectangle ABCD is formed in a circle as shown in Figure. If AE = 8 cm and AD = 5 cm, find the perimeter of the rectangle. (NCERT Exemplar)

Solution:

DE (Radius) = AE + AD = 8 cm + 5 cm = 13 cm

DB = AC = 13 cm (Diagonal of a rectangle are equal)

In right âˆ†ADC,

AD^{2} + DC^{2} = AC^{2} (By Pythagoras Theorem)

â‡’ (5)^{2} + DC^{2} = (13)^{2}

â‡’ 25 + DC^{2} = 169

â‡’ DC^{2} = 169 – 25 = 144

â‡’ DC = âˆš144 = 12 cm

Perimeter of rectangle ABCD = 2(AD + DC)

= 2(5 cm + 12 cm)

= 2 Ã— 17 cm

= 34 cm

Question 19.

Find the area of a parallelogram-shaped shaded region. Also, find the area of each triangle. What is the ratio of the area of shaded portion to the remaining area of the rectangle?

Solution:

Here, AB = 10 cm

AF = 4 cm

FB = 10 cm – 4 cm = 6 cm

Area of the parallelogram = Base Ã— Height = FB Ã— AD = 6 cm Ã— 6 cm = 36 cm^{2}

Hence, the required area of shaded region = 36 cm^{2}.

Area âˆ†DEF = \(\frac { 1 }{ 2 }\) Ã— b Ã— h

= \(\frac { 1 }{ 2 }\) Ã— AF Ã— AD

= \(\frac { 1 }{ 2 }\) Ã— 4 Ã— 6

= 12 cm^{2}

Area âˆ†BEC = \(\frac { 1 }{ 2 }\) Ã— b Ã— h

= \(\frac { 1 }{ 2 }\) Ã— GC Ã— BC

= \(\frac { 1 }{ 2 }\) Ã— 4 Ã— 6

= 12 cm^{2}

Area of Rectangle ABCD = l Ã— b = 10 cm Ã— 6 cm = 60 cm^{2}

Remaining area of Rectangle = 60 cm^{2} – 36 cm^{2} = 24 cm^{2}

Required Ratio = 36 : 24 = 3 : 2

Question 20.

A rectangular piece of dimension 3 cm Ã— 2 cm was cut from a rectangular sheet of paper of dimensions 6 cm Ã— 5 cm. Find the ratio of the areas of the two rectangles. (NCER T Exemplar)

Solution:

Length of the rectangular piece = 6 cm

Breadth = 5 cm

Area of the sheet = l Ã— b = 6 cm Ã— 5 cm = 30 cm^{2}

Area of the smaller rectangular piece = 3 cm Ã— 2 cm = 6 cm^{2}

Ratio of areas of two rectangles = 30 cm^{2} : 6 cm^{2} = 5 : 1

### Perimeter and Area Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 21.

In the given figure, ABCD is a square of side 14 cm. Find the area of the shaded region.

(Take Ï€ = \(\frac { 22 }{ 7 }\))

Solution:

PQ = \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) Ã— 14 = 7 cm

PQRS is a square with each side 7 cm

Radius of each circle = \(\frac { 7 }{ 2 }\) cm

Area of the quadrants of each circle = \(\frac { 1 }{ 4 }\) Ã— Ï€r^{2}

Area of the four quadrants of all circles

Area of the square PQRS = Side Ã— Side = 7 cm Ã— 7 cm = 49 cm^{2}

Area of the shaded portion = 49 cm^{2} – 38.5 cm^{2} = 10.5 cm^{2}

Hence, the required area = 10.5 cm^{2}.

Question 22.

Find the area of the following polygon if AB = 12 cm, AC = 2.4 cm, CE = 6 cm, AD = 4.8 cm, CF = GE = 3.6 cm, DH = 2.4 cm.

Solution:

BE = AB – AE

= 12 cm – (AC + CE)

= 12 cm – (2.4 cm + 6 cm)

= 12 cm – 8.4 cm

= 3.6 cm

Area of the polygon AFGBH = Area of âˆ†ACF + Area of rectangle FCEG + Area of âˆ†GEB + Area of âˆ†ABH

= 3.6 cm^{2} + 4.32 cm^{2} + 21.6 cm^{2} + 6.48 cm^{2} + 14.4 cm^{2}

= 50.40 cm^{2}

Hence, the required area = 50.40 cm^{2}.