## NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 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 m²

(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

Side of the square = Perimeter/4

= 320/4 m = 80m

Area of the square = Side × Side

= 80 m × 80 m = 6400 m²

**Ex 11.1 Class 7 Maths Question 3.**

**Find the breadth of a rectangular plot of land, if its area is 440 m² and the length is 22 m. Also, find its perimeter.**

**Solution:**

Given: Area = 440 m²

Length = 22 m

Breadth = Area/Length = 440m²/22m = 20m

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 cm²

**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= 60m × 60m/90

⇒ 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 cm2

Area of square shape = (Side)2

= (31)2 = 961 cm2

Since 961 cm2 > 880 cm2

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 cm²

**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 m².**

**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 m²

= 16.20 m²

Area of the door = l × b = 2 m × 1 m = 2 m²

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

= 16.20 m² – 2 m²= 14.20 m²

Cost of white washing

= ₹ 14.20 × 20 = ₹ 284.00

Hence, the required area = 14.20 m2 and the required cost = ₹ 284