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.
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
Blazeš„
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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