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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

Sagot :

Answer: [tex]75\ ft^2[/tex]

Step-by-step explanation:

Given

The dimension of playspace is [tex]10\ ft\times 12\ ft[/tex]

If the length and width is increased by 3 ft

New length is  [tex]13\ ft[/tex]

New width is [tex]15\ ft[/tex]

So, the dimension becomes [tex]13\ ft\times 15\ ft[/tex]

The difference between the areas of the playspace is

[tex]\Rightarrow 13\times 15-10\times 12\\\Rightarrow 195-120=75\ ft^2[/tex]

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