Question 102369: The hypotonuse of a right triangle is 3m more than twice the short leg. The longer leg is 2 m more than twice the shorter leg. Find the lengths of the sides.
This is what I have so far, but am stuck.
x^2+(2x+2)^2=(2x+3)^2
x^2+4x^2+4x+4x+4 = 4x^2+6x+6x+9
x^2+4x^2+8x+4=4x^2+12x^2+9
now what??
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! Let s=shorter leg, L=longer leg, h=hypotenuse
h=2s+3
L=2s+2
s^2+(2s+2)^2=(2s+3)^2
s^2+4s^2+8s+4=4s^2+12s+9
s^2-4s-5=0 subtract the right side from the left side.
(s-5)(s+1)=0 Factor.
s=-1, s=5
s can't equal -1 so s=5 is the only answer.
s=5, L=12, h=13
Ed
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