Asked by Jonel

One side of a triangle is 5cm more than the other and 2cm less than the hypotenuse. Find the lengths of the 3 sides?
Answered by Reiny
let the shortest side be x
then the longer side is x+5 and the
hypotenuse is x+7

(x+7)^2 = x^2 + (x+5)^2
x^2 + 14x + 49 = x^2 + x^2 + 10x + 25
x^2 - 4x - 24 = 0
x^2 - 4x + .... = 24 + ... , going to complete the square
x^2 - 4x + 4 = 24+4 = 28
(x+2)^2 = 28
x-2 = ± √28
x = ±√28 + 2
can't have a negative side, so x = √28 - 2


the sides are √28 + 2 , √28 + 7, and √28+9
or appr 7.29 , 12.29, and 14.29

check:
14.29^2 = 204.20
12.29^2 + 7.29^2 = 204.19 , close enough for 2 decimal accuracy
Answered by Reiny
the third last line of the solution should of course be
<b>(x-2)^2 = 28 </b>
and the conclusion line:
<b>can't have a negative side, so x = √28 + 2</b>
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