Asked by Anonymous
The perimeter of a rectangle is
72 feet. The length of the rectangle is twice the width. Find the length and width of the rectangle.
72 feet. The length of the rectangle is twice the width. Find the length and width of the rectangle.
Answers
Answered by
Kolb
So if P=72ft , let's find an equation to compare! Perimeter=(Length x 2)+(width x 2) but we know that the length is twice the width or 2L=w , so let's substitute this back into our equation, P=(2L)(2w)=(2L)(2(2L))=8L^2=72
L^2=9
L=+-3 but it doesn't make sense to have a negative length so L must be 3. Plugging this back in, we get that 2L=w so w=6
L^2=9
L=+-3 but it doesn't make sense to have a negative length so L must be 3. Plugging this back in, we get that 2L=w so w=6
Answered by
Reiny
L = 2W
2L + 2W = 72
L + W = 36
2W + W = 36
3W = 36
W = 12 ----> L = 24
The rectangle is 24 ft long and 12 feet wide
check:
perimeter = 24+24+12+12 = 72
2L + 2W = 72
L + W = 36
2W + W = 36
3W = 36
W = 12 ----> L = 24
The rectangle is 24 ft long and 12 feet wide
check:
perimeter = 24+24+12+12 = 72
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