Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 44 Answers Solutions Chapter 12 Perimeter and Area.

## Perimeter and Area Class 7 Practice Set 44 Answers Solutions Chapter 12

Question 1.
If the length and breadth of a rectangle are doubled, how many times the perimeter of the old rectangle will that of the new rectangle be?
Solution:
Let the length of the old rectangle be l and breadth be b.
∴ Perimeter of old rectangle = 2(l + b)
Length of new rectangle = 2l and breadth = 2b
∴ Perimeter of new rectangle = 2(2l + 2b)
= 2 x 2 (l + b)
= 2 x perimeter of old rectangle
∴ The perimeter of new rectangle will be twice the perimeter of old rectangle.

Question 2.
If the side of a square is tripled, how many times the perimeter of the first square will that of the new square be?
Solution:
Let the length of the square be a.
Perimeter of square = 4 x side
= 4 x a = 4a
Side of new square = 3 x a = 3a
Perimeter of new square = 4 x side
= 4 x 3a = 3 x 4a = 3x perimeter of original square.
∴ The perimeter of new square will be three times the perimeter of original square.

Question 3.
Given alongside is the diagram of a playground. It shows the length of its sides. Find the perimeter of the playground. Solution: Side AF = side BC + side DE
∴ Side AF = 15 + 15 = 30 m
Side FE = side AB + side CD
∴ Side FE = 10 + 5 = 15 m
∴ Perimeter of the playground = side AB + side BC + side CD + side DE + side FE + side AF
= 10 + 15 + 5 + 15 + 15 + 30
= 90 m.
∴ The perimeter of the playground is 90 m.

Question 4.
As shown in the figure, four napkins all of the same size were made from a square piece of cloth of length 1 m. What length of lace will be required to trim all four sides of all the napkins? Solution:
Side of the square piece of cloth = 1 m
∴ Side of each napkin = 0.5 m
Length of lace that will be required for 1 napkin = perimeter of the napkin
= 4 x side = 4 x 0.5 = 2 m
∴ Perimeter of 4 napkins = 4 x 2 = 8 m
∴ 8 metre long lace will be required to trim all four napkins.