The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle. Define variables, set up equation and solve.
2007-01-21 16:45:19 · 2 answers · asked by sittinncat 1 in Science & Mathematics ➔ Mathematics
In a word problem, always assign the variable to what everything is being compared to. In this case, it's the width.
Let w = the width of the rectangle. Then the length would be
w - 2.
The perimeter of a rectangle is given by the following formula.
P = (2 times the length) + (2 times the width)
But we know the perimeter, the length, and the width.
52 = 2L + 2W
52 = 2(w - 2) + 2(w)
52 = 2w - 4 + 2w
52 = 4w - 4
Adding 4 to both sides,
56 = 4w
Therefore, w = 14. Since the length is w - 2, the length is 12.
So the dimension of the rectangle are 14 x 12.
2007-01-21 16:52:30 · answer #1 · answered by Puggy 7 · 0⤊ 0⤋
l=2w+2
2w=l-2
p=52=2l+2w=2l+l-2
52=3l-2
3l=54
l=18
2w=l-2=18-2=16
w=8
2007-01-22 00:52:36 · answer #2 · answered by yupchagee 7 · 0⤊ 0⤋