SOLUTION: if the length of a rectangle measure measure is twice as much as a width and the perimeter is 72 feet find the dimensions of the rectangle
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Question 950262: if the length of a rectangle measure measure is twice as much as a width and the perimeter is 72 feet find the dimensions of the rectangle
Found 2 solutions by jhaashok630, macston:
Answer by jhaashok630(18) (Show Source): You can put this solution on YOUR website!
length of a rectangle measure measure is twice as much as a width and the perimeter is 72 feet
Let x=width of rectangle
length = 2x
Perimeter = 72
2l+2w=72
4x+2x=72
x=12
length = 24 feet and width = 12 feet
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
W=width; L=length=2W; P=perimeter=2(L+W)=72 feet
P=2(L+W) Substitute for L
72 feet=2(2W+W) Divide each side by 2.
36 feet=3W Divide each side by 3.
12 feet=W ANSWER 1: The width is 12 feet.
L=2w=2(12 ft)=24 ft ANSWER 2: The length is 24 feet.
CHECK
P=2(L+W)
72 feet=2(24 feet+12 feet)
72 feet=2(36 feet)
72 feet=72 feet
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