SOLUTION: The lenght of a rectangle is one inch less than twice its width. The diagonal of the rectangle is two inches more than its length. Find the area of the rectangle.
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Question 101992: The lenght of a rectangle is one inch less than twice its width. The diagonal of the rectangle is two inches more than its length. Find the area of the rectangle. Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! L=2W-1
c=diagonal and is the hypotenuse of a triangle with sides L and W
c=2W+1
a^2+b^2=c^2
(2W-1)^2+W^2=(2W+1)^2
4W^2-4W+1+W^2=4W^2+4W+1
W^2-8W=0
W(W-8)=0
W=0, W=8 Since the rectangle can't have a width of 0 W=0 is not an answer.
W=8 is the only answer.
L=15
c=17
A=15*8=120sq inches (Answer)
check:
8^2+15^2=64+225=289
17^2=289
Ed
