SOLUTION: The length of a rectangle is 2 in. less than four times the width. If the diagnol is 4 in. more than the length, find the dimensions of the rectangle.
The width of the rectangle
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Question 634415: The length of a rectangle is 2 in. less than four times the width. If the diagnol is 4 in. more than the length, find the dimensions of the rectangle.
The width of the rectangle is ?
The length of the rectangle is ?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
The length of a rectangle is 2 in. less than four times the width. If the diagnol is 4 in. more than the length, find the dimensions of the rectangle.
The width of the rectangle is ?
The length of the rectangle is ?
**
let w=width
4w-2=length
length of diagonal=4w-2+4=4w+2
by pythagorean theorem:
diagonal^2=width^2+length^2
(4w+2)^2=w^2+(4w-2)^2
16w^2+16w+4=w^2+16w^2-16w+4
w^2-32w=0
w(w-32)=0
w=0 (reject, w>0)
or
w=32
4w-2=128-2=126
4w+2=128+2=130
..
width of the rectangle is=32 in
length of the rectangle is=126 in
length of diagonal=130 in
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