Question

The figure below represents a play space that Logan fenced in for his dog.

10 ft
- 12 ft2
Logan is getting a second dog and wants to increase the length of the play
space by 3 feet and the width by 3 feet. What will be the difference in the area,
in square feet, between the original play space and the new play space?
Show your work.

Answer 1

Difference in area = 75 ft²

Explanation:

Case 1 (When logan had only one dog):

length of the play space=
`12` ft

breadth of the play space=
`10` ft

we know that area of a rectangle is length times breadth (length x breadth)

therefore, area of the play space=
`12*10`

=
`120ft^(2)`

Case 2(When Logan had two dogs):

Given: Logan wants to increase the length of the play space by 3 feet and width by 3 feet

⇒length of the play space=
`12+3` ft=
`15` ft

breadth of the play space=
`10+3` ft=
`13` ft

we know that area of a rectangle is length times breadth (length x breadth)

therefore, area of the play space=
`15*13`

=
`195ft^(2)`

Therefore difference in the area of play space =
`195-120`

=75 ft²