Asked by samantha

a rectangle is twice as long as it is wide. if both of its dimensions are increased by 4m, its area is increased by 88m^2 find the dimensions of the original rectangle

Original rectangle = w for width and 2w for length. Area = w x 2w

Larger rectangle = w+4 for width and 2w+4 for length. Area = (w+4)*(2w+4)

The problem states that the larger rectangle has an area that is 88 m<sup>2</sup> more than the original.
Put that in equation form.
area of rectangle 2 - area of rectangle 1 = 88.

(w+4)*(2w+4)-(w*2w)=88.

Solve for w. The length will be twice that. Post your work if you run into trouble.

Answers

Answered by Deez nuts
I'm still so confused wtf
Answered by clanther
36m
Answered by Cynthia Valdez
18
Answered by BlazešŸ”„
Deez Nuts I don't know if I don't think I can help but I use Algebra . com. (I set the url up weird because urls are not allowed to be sent)
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