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 m2 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.
Expanding the left side of the equation, we get:
2w^2 + 8w + 4w + 16 - 2w^2 = 88
Combining like terms, we have:
12w + 16 = 88
Subtracting 16 from both sides, we get:
12w = 72
Dividing both sides by 12, we find:
w = 6
The width of the original rectangle is 6m, and the length is twice that, so the length is 12m. Therefore, the dimensions of the original rectangle are 6m by 12m.