## MP Board Class 7th Maths Solutions Chapter 11 Perimeter and Area Ex 11.1

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 m^{2} of the land costs ₹ 10,000.

Solution:

Length (l) = 500 m

Breadth (b) = 300 m

(i) Area = Length × Breadth = 500 × 300 = 150000 m^{2}

(ii) Cost of 1 m^{2} land = ₹ 10000

∴ Cost of 150000 m^{2} land

= 150000 × 10000 = ₹ 1500000000

Question 2.

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

Solution:

Perimeter of the square park = 320 m

∴ 4 × Length of the side of park = 320

Length of the side of park = \(\frac{320}{4}\) = 80 m

Area = (Length of the side of park)^{2}

= (80)^{2} = 6400 m^{2}

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:

Area of a rectangular plot = 440 m^{2}

Length = 22 m

Area = Length x Breadth = 440 m^{2}

∴ 22 × Breadth = 440

⇒ Breadth = \(\frac{440}{22}\) = 20 m

∴ Perimeter = 2 (Length + Breadth)

= 2 (22 + 20) = 2(42) = 84 m

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:

Length = 35 cm

Perimeter = 100 cm

∴ 2 (35 + Breadth) = 100

⇒ 35 + Breadth = 50

⇒ Breadth = 50 – 35 = 15 cm

∴ Area = Length × Breadth

= 35 × 15 = 525 cm^{2}

Question 5.

The area of a square park is the 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:

Side of the square park = 60 m

Length of the rectangular park = 90 m

Area of the square park = (side)^{2} = (60)^{2} = 3600 m^{2}

Area of rectangular park = Length × Breadth

= 90 × Breadth

It is given that area of square park = area of rectangular park

∴ 3600 = 90 × Breadth

⇒ Breadth = 40 m

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:

Length of rectangle = 40 cm

Breadth of rectangle = 22 cm

Perimeter of rectangle = Perimeter of square

∴ 2 (Length + Breadth) = 4 × Side of square

⇒ 2 (40 + 22) = 4 × Side of square

⇒ 2 × 62 = 4 × Side of square

∴ Side of square = \(\frac{124}{4}\) = 31 cm

Now, area of rectangle = 40 × 22 = 880 cm^{2}

Area of square = (Side)^{2} = 31 × 31 = 961 cm^{2}

As 961 > 880.

Therefore, the square-shaped wire encloses more area than rectangle – shaped wire.

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:

Breadth = 30 cm

Perimeter = 130 cm

∴ 2 (Length + 30) = 130

⇒ Length + 30 = 65

⇒ Length = 65 – 30 = 35 cm

Now, area = Length × Breadth

= 35 × 30 = 1050 cm^{2}

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 (see the given figure). Find the cost of white washing the wall, if the rate of white washing the wall is ₹ 20 per m^{2}.

Solution:

Length of wall = 4.5 m

Breadth of wall = 3.6 m

Area of wall = Length × Breadth

= 4.5 × 3.6

= 16.2 m^{2}

Area of door = 2 × 1 = 2 m^{2}

Area to be white-washed

= Area of wall – Area of door

= 16.2 – 2 = 14.2 m^{2}

Cost of white-washing 1 m^{2} area = ₹ 20 2.

∴ Cost of white-washing 14.2 m^{2} area

= 14.2 × 20 = ₹ 284