Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths 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 mĀ² of the land costs ā¹ 10,000.

Solution:

(i) Length of the rectangle (I) = 500 m

Breadth of the rectangle (b) = 300 m

ā“ Area = l x b

= 500 m x 300 m = 150000 mĀ²

(ii) āµ Cost of 1 mĀ² of land = ā¹ 10,000

ā“ Cost of 150000 mĀ² of land

= ā¹ 10000 x 150000 = ā¹ 1,50,00,00,000

Question 2.

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

Solution:

āµ Perimeter of the square park = 320 m

i.e. 4 x Side = 320 m

ā“ Side = \(\frac { 320 }{ 4 }\) = m = 80 m

Now, area of the square park = Side x Side = 80 m x 80 m = 6400 mĀ²

Question 3.

Find the breadth of a rectangular plot of land, if its area is 440 mĀ² and the length is 22 m. Also find its perimeter.

Solution:

Length of the rectangle = 22 m

Suppose breadth of rectangle = b m

ā“ Area of the rectangle = 22 x b

or 22 x b = 440 (āµ Area = 440 m2)

or b = \(\frac { 440 }{ 22 }\) = 20 m

Thus, the breadth of the rectangle = 20m

Now, perimeter of a rectangle = 2(1 + b)

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

Perimeter of the rectangular sheet = 100 cm

Length (l) = 35 cm, Suppose breadth = b cm

ā“ 2(1 + b) = 100 cm

or 2(35 + b) = 100 cm

or 35 + b = \(\frac { 100 }{ 2 }\) cm = 50 cm

or b = (50 – 35) cm = 15 cm

ā“ Breadth of the rectangular sheet =15 cm

Now, area of a rectangle = l x b

ā“ Area of the rectangular sheet = 35 x 15 cmĀ² = 525 cmĀ²

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

ā“ Area of the square park = 60 x 60 mĀ²

= 3600 mĀ²

āµ [Area of the rectangular park] = [Area of the square park] 3600 mĀ²

āµ Length of the rectangular park = 90 m

ā“ Breadth of the rectangular park

= \(\frac { 3600 }{ 90 }\) m = 40 m.

Question 6.

A wire is in the shape of a rectangle. Its length is 40 cm arid 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 (l) = 40 cm, Breadth (b) = 22 cm

ā“ Perimeter = 2(l + b) = 2(40 + 22) cm

= 2(62) cm = 124 cm

ā“ Area of the rectangle = l x b

= 40 cm x 22 cm = 880 cmĀ²

āµ The same wire is rebent to form a square.

ā“ Perimeter of square = Perimeter of the rectangle

ā“ Perimeter of the square = 124 cm

ā“ Side of the square = \(\frac { 124 }{ 4 }\) = 31 cm

Area of the square = 31 cm x 31 cm = 961 cmĀ²

āµ 961 cmĀ² > 880 cmĀ²

ā“ Area of the square is more than the rectangle.

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:

Perimeter of the rectangle = 130 cm

Breadth of the rectangle = 30 cm

Let the length of the rectangle be ‘l’.

ā“ 2(1 + 30) = 130

or l + 30 = \(\frac { 130 }{ 2 }\) = 65

or l = 65 – 30 = 35 cm

ā“ Length of the rectangle = 35 cm

Now, area of the rectangle = l x b

= 35 x 30 cmĀ² = 1050 cmĀ²

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 ā¹ 20 per mĀ².

Solution:

Length of the wall (l) = 4.5 m

Breadth of the wall (b) = 3.6 m

ā“ Area of the wall = l x b

= (4.5 m) x (3.6 m)

= \(\frac { 45 }{ 10 }\) x \(\frac { 36 }{ 10 }\)

= 16.2 mĀ²

Area of the door = 2m x 1 m = 2mĀ²

ā“ Area to be white washed

= [Area of the wall] – [Area of the door]

= 16.2 mĀ² – 2 mĀ² = 14.2 mĀ²

Since, rate of white washing = ā¹ 20 per mĀ²

ā“ Cost of white washing = ā¹ 20 x (14.2) = ā¹ 284