The length of a rectangle is 1 in. more than twice its width. If the
perimeter of the rectangle is 74 in., find the dimensions of the rectangle.
2007-05-30 09:32:13 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You are given:
L = 2W + 1
P = 74
But perimeter of a rectangle is twice the length plus twice the width (2L + 2W)... so we have:
2L + 2W = 74
Substitute the first equation (replace L with 2W+1) into the second:
2L + 2W = 74
2(2W+1) + 2W = 74
4W + 2 + 2W = 74
6W = 72
W = 12
Going back to the original equation, and using the now-known value of W, you can solve for L:
L = 2W + 1
L = 2*12 + 1
L = 25
We have width=12 and length=25. Those are the dimensions that you're asked to calculate.
Now, let's check and make sure the perimeter works out:
2L + 2W =? 74
2*25 + 2*12 =? 74
50 + 24 =? 74
74 = 74
2007-05-30 09:37:08 · answer #1 · answered by McFate 7 · 0⤊ 0⤋
It is 12 in wide and 25 in long.
You get this by solving this :
Width = x and length = 2x + 1
There are two of each in a perimeter.
2(x) + 2(2x + 1) = 74
2x + 4x + 2 = 74
6x = 72
x = 12
so 2x + 1 = 2(12) + 1 = 25
2007-05-30 16:44:54 · answer #2 · answered by Don E Knows 6 · 0⤊ 0⤋
Length = 2w + 1
Perimeter = 2w + 2(2w + 1) = 74
2w + 4w+ 2 = 74
6w = 72
width = 12
Length = 2w + 1 = 25
Area = length * width
Area = 25 * 12 = 300 in²
2007-05-30 16:43:13 · answer #3 · answered by Robert L 7 · 0⤊ 0⤋
12" x 25" to break down what he said.
2007-05-30 16:40:51 · answer #4 · answered by Randi 2 · 0⤊ 0⤋