The length of a rectangle is twice its width. If the area of the rectangle is 162 yds ^2, find its perimeter.
2006-12-03 06:22:51 · 5 answers · asked by Ken H 1 in Science & Mathematics ➔ Mathematics
Based on the given information, you can set up an equation
Let X = width Let 2X = length
2X * X = 162
2X^2=162
X^2=81
X=9
So the dimensions are 9 and 18
Thus the perimeter is 54 yds
2006-12-03 06:27:35 · answer #1 · answered by Panky1414 2 · 0⤊ 0⤋
A = LW
162 = 2w(w)
162 = 2w²
162 / 2 = 2w²/2
81 = w²
â81 = w²
9 = w
The answer is w = 9
The width = 9 yards
the length = two times the width = 18
- - - - - - - - - - -
Find the perimeter
Perimeter Formula
P = 2L + 2W
P = 2(18) + 2(9)
P = 26 + 18
P = 54 Yards
The answer is P = 54 yards
- - - - - - - s-
2006-12-03 06:58:58 · answer #2 · answered by SAMUEL D 7 · 0⤊ 0⤋
area=LxW
L=2x
W=x
(2x)(x)=162 sq yds
2x^2=162
x^2=81
x=9 is the width
Therefore, if the width is 9 and the length is twice that, the length is 18.
2006-12-03 06:29:14 · answer #3 · answered by Tony T 4 · 0⤊ 0⤋
54
2006-12-03 06:30:47 · answer #4 · answered by Anonymous · 0⤊ 0⤋
L=2W
L.W=162
2(L+W)=?
Hence 2 W.W=162
W.W=81
W=9
L=2.9=18
2(18+9)=2.27= 54
2006-12-03 06:28:04 · answer #5 · answered by iyiogrenci 6 · 0⤊ 0⤋