Question 89945This question is from textbook
: Could you please help me with this problem?:The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm, what are the dimensions (length and the width) of the rectangle? Thank you.
This question is from textbook
Answer by malakumar_kos@yahoo.com(315)   (Show Source):
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The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm, what are the dimensions (length and the width) of the rectangle
Let the width of the rectangle be = w = x cm
the length of the rectangle be = l = (x+1) cm
diagonal of the rectangle = d = 4 cm
therefore diagonal^2 = l^2+w^2 (use pythagoras theorem)
4^2 = (x+1)^2+x^2
16 = x^2+2x+1+x^2
2x^2+2x+1-16 = 0
2x^2+2x-15 = 0
compare with the std form , a=2 ,b = 2 &c = -15
x = -2+-sq rt 2^2-4*2*-15/2*2
= -2+-sq rt 4+120/4
= -2+-sq rt 124/4
= 2(-1+-sq rt 31)/4
(-1+-sq rt 31)/2
width of the rectangle = w = (-1+-sq rt 31)/2 cm
length of th rectangle = l = 1+(-1+-sq rt 31)/2 cm
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