NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area

### NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1

**NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.1**

Ex 11.1 Class 7 Maths Question 1.

The length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. Find

(i) its area

(ii) the cost of the land, if 1 m2 of the land costs ₹ 10,000

Solution:

Given: l = 500 m, b = 300 m

(i) Area = l × b

= 500 m × 300 m = 150000 m^{2}

(ii) Cost of land = ₹ 10,000 × 150000 = ₹ 15,00,000,000

Ex 11.1 Class 7 Maths Question 2.

Find the area of a square park whose perimeter is 320 m.

Solution:

Given: Perimeter = 320 m

Area of the square = Side × Side

= 80 m × 80 m = 6400 m^{2}

Ex 11.1 Class 7 Maths Question 3.

Find the breadth of a rectangular plot of land, if its area is 440 m^{2} and the length is 22 m. Also, find its perimeter.

Solution:

Given: Area = 440 m^{2}

Length = 22 m

Perimeter = 2[l + b] = 2 [22 m + 20 m]

= 2 × 42 m = 84 m

Ex 11.1 Class 7 Maths Question 4.

The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.

Solution:

Given: Perimeter = 100 cm

Length = 35 cm

Perimeter = 2(l + b)

100 = 2(35 + b)

100/2=35+b

⇒ 50 = 35 + b

⇒ b = 50 – 35 = 15 cm

∴ Breadth = 15 cm

Area = l × b = 35 cm × 15 cm

= 525 cm^{2}

Ex 11.1 Class 7 Maths Question 5.

The area of a square park is same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.

Solution:

Given: Side of the square park = 60 m Length of the rectangular park = 90 m Area of the rectangular park = Area of the square park

90 m × 6 = 60 m × 60 m

⇒ b = 40m

Hence, the required breadth = 40 m.

Ex 11.1 Class 7 Maths Question 6.

A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side. Also find which shape encloses more area?

Solution:

Given: Length = 40 cm, Breadth = 22 cm Perimeter of the rectangle

= Length of the wire

= 2(l + b) = 2(40 cm + 22 cm)

= 2 × 62 cm = 124 cm

Now, the wire is rebent into a square.

Perimeter = 124 cm

⇒ 4 × side = 124

∴ side = 124/4 cm = 31 cm

So, the measure of each side = 31 cm

Area of rectangular shape = l × b

= 40 cm x 22 cm

= 880 cm^{2}

Area of square shape = (Side)^{2}

= (31)^{2} = 961 cm^{2}

Since 961 cm^{2} > 880 cm^{2}

Hence, the square encloses more area.

Ex 11.1 Class 7 Maths Question 7.

The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also find the area of the rectangle.

Solution:

Given: Perimeter = 130 cm

Breadth = 30 cm

Perimeter = 2 (l + b)

130 cm = 2(l + 30 cm)

⇒ 130/2 cm = l + 30 cm

⇒ 65 cm = l + 30 cm

⇒ 65 cm – 30 cm = l

∴ l = 35 cm

Area of the rectangle = l × b = 35 cm × 30 cm

= 1050 cm^{2}

Ex 11.1 Class 7 Maths Question 8.

A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m. Find the cost of white washing the wall, if the rate of white washing the wall is ₹ 20 per m^{2}.

Solution:

Given: Length of wall = 4.5 m

Breadth of the wall = 3.6 m

Length of the door = 2 m

Breadth of the door = 1 m

Area of the wall = l × b = 4.5 m × 3.6 m = 16.20 m^{2}

= 16.20 m^{2}

Area of the door = l × b = 2 m × 1 m = 2 m^{2}

∴ Area of the wall to be white washed = Area of the wall – Area of the door

= 16.20 m^{2} – 2 m^{2}= 14.20 m^{2}

Cost of white washing

= ₹ 14.20 × 20 = ₹ 284.00

Hence, the required area = 14.20 m^{2} and the required cost = ₹ 284

**Exercise 11.2**

**Exercise 11.3**

**Exercise 11.4**