The length of a rectangle exceeds its width by 4 feet. If the width is doubled and the length is diminished by 2 feet, a new rectangle is formed whose perimeter 8 feetmore than the perimeter of the actual rectangle. Find the dimensions of the original rectangle.
2007-01-07 10:09:05 · 1 answers · asked by Rani 1 in Science & Mathematics ➔ Mathematics
Let
L = length of original rectangle
W = width of original rectangle
P = perimeter of original rectangle
M = length of new rectangle
X = width of new rectangle
Q = perimeter of new rectangle
We have
L = W + 4
P = 2L + 2W = 2(W + 4) + 2W = 4W + 8
M = 2W
X = W - 2
Q = 2M + 2X = 2(2W) + 2(W - 2) = 4W + 2W - 4 = 6W - 4
Q = P + 8
6W - 4 = (4W + 8) + 8 = 4W + 16
2W = 20
W = 10
L = W + 4 = 10 + 4 = 14
The original rectangle was 10 feet by 14 feet.
2007-01-07 10:51:19 · answer #1 · answered by Northstar 7 · 1⤊ 0⤋