Question

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:

$(3x+3y+9)\ ft^{2}$

Step-by-step explanation:

Let

x------> the length of the original play space

y-----> the width of the original play space

we know that

The area of the original play space is equal to

$A=xy$

The area of the new play space is equal to

$An=(x+3)(y+3)\\An=xy+3x+3y+9$

Find the difference

$An-A=(xy+3x+3y+9)-(xy)\\An-A= (3x+3y+9)\ ft^{2}$