The length of a rectangle is 2 yd longer than its width. If the perimeter of the rectangle is 28 yd , find its area.
2007-07-10 14:59:42 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
L = W + 2
2W + 2L = 28
Substitute first (replace L with W+2) into the second:
2W + 2(W+2) = 28
4W + 4 = 28
4W = 24
W = 6
L = W + 2 = 6 + 2 = 8
Knowing the width and length, we can compute area:
A = L * W
A = 8 * 6
A = 48 square yd
2007-07-10 15:03:27 · answer #1 · answered by McFate 7 · 0⤊ 2⤋
easy
Width of the Rectangle be = x yds
Its length = x + 2 yds
The perimeter of the rectangle = 2 ( l + b)
= 2 (x + x + 2)
4x + 4 = 28 yds
=> 4x = 24 yds
=> x = 24 / 4 = 6 yds.
Length + 6+2 = 8 yds
Width = 6 yds
Area = l * b = 8 8 6 = 48 sq. yds. Answer
..
2007-07-10 22:05:20 · answer #2 · answered by Anonymous · 0⤊ 1⤋
2 sides are 6 long and the other 2 sides are 8 long
Area is 48
2007-07-10 22:03:35 · answer #3 · answered by Nick L 1 · 0⤊ 1⤋
w = width
L = length
Perimeter = 2L + 2w
Area = L*w
L = w + 2
2L + 2w = 28
Substitute and solve:
2(w+2) + 2w = 28
2w + 4 + 2w = 28
4w = 24
w = 6
L = w + 2
L = (6) + 2
L = 8
A = L*w = 6*8 = 48 yd^2
2007-07-10 22:02:45 · answer #4 · answered by whitesox09 7 · 1⤊ 2⤋
Oh, goody, I love riddles!
Hey, wait a second... this just looks like some lame homework problem. What a gyp!
2007-07-10 22:25:40 · answer #5 · answered by Anonymous · 0⤊ 1⤋
trial and error
2007-07-10 22:07:11 · answer #6 · answered by abster_x0x 2 · 0⤊ 1⤋