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Question 127444: Hello.
This is my first time using this site. I hope I've submitted this question correctly. My problem is:
The hypotenuse of a right triangle is 28 inches longer than the shortest side, and 14 inches longer than the remaining. Find the dimensions of the triangle.
I tried this so far:
the hypotenuse(a) is x
the shortest side(b) is x-28
the remaining side(c) is x-14
P=a+b+c
P= x+(x-28)+(x-14)
3x-28-14
3x-42
x=14
i don't think i'm onm the right track at all... can you help?
thank you in advance...
maurice
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! yes, you are off the track
this is a Pythagorean problem __ b^2+c^2=a^2
let x=hypotenuse __ (x-28)^2+(x-14)^2=x^2 __ x^2-56x+784+x^2-28x+196=x^2
x^2-84x+980=0
factoring __ (x-70)(x-14)=0
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