Question
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
I'm still so confused wtf
36m
18
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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