one leg of a right triangle is 7 feet shorter then then other leg. the hypotenuse is 2 feet longer then the longer leg. find the lengths of all three sides of the right triangle. this shorter then is throwing me for a loop. somebody help me .please

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- Apr 23rd 2005, 03:07 PM #1RSPRAGGSGuest
## [SOLVED] help

one leg of a right triangle is 7 feet shorter then then other leg. the hypotenuse is 2 feet longer then the longer leg. find the lengths of all three sides of the right triangle. this shorter then is throwing me for a loop. somebody help me .please

- Apr 23rd 2005, 03:20 PM #2
## I think this is it

I think this is how you do it, somebody correct me if I'm wrong.

one leg of a right triangle is 7 feet shorter then then other leg. => x-7

the hypotenuse is 2 feet longer then the longer leg => x+2

So the two sides are x, x-7, and the hypotenuse is x+2

By Pathagores theorum we know that (a^2)+(b^2)=(c^2)

So we have:

(x^2)+(x-7)^2 = (x+2)^2

we then get:

x^2 + x^2 -14x +49 = x^2 +4x +4

Simplifies to:

x^2 -18x +45 = 0

We factor to get:

(x-3)(x-15) = 0

So x can equal 3 or 15

Since one side is 7 less than x and we cant have negative length 3 cannot be an answer. So one side is 15, the other is 8, and the hypotonus is 17.