Question

The play space is 10 ft to 12 ft. 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

Answer 1

Answer: $75\ ft^2$

Step-by-step explanation:

Given

The dimension of playspace is $10\ ft\times 12\ ft$

If the length and width is increased by 3 ft

New length is  $13\ ft$

New width is $15\ ft$

So, the dimension becomes $13\ ft\times 15\ ft$

The difference between the areas of the playspace is

$\Rightarrow 13\times 15-10\times 12\\\Rightarrow 195-120=75\ ft^2$

Answer 2

Answer: 75 square feet

Step-by-step explanation:

The original dimensions are 10 ft by 12 ft, so the area is (10 f)(15 f)