SOLUTION: i have another ? The longer leg of a right triangle has length 1cm less than twice the shorter leg. the hypotenuse has length 1cm greater than the shorter leg. find the lengths

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Question 119171This question is from textbook
: i have another ?
The longer leg of a right triangle has length 1cm less than twice the shorter leg. the hypotenuse has length 1cm greater than the shorter leg. find the lengths of the three sides of the triangle.
i have:
x
1-x
2x-1 This question is from textbook

Found 2 solutions by solver91311, josmiceli:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
I presume x to be the short leg, so your 2x - 1 representation would be correct for the longer leg, but the hypotenuse is then x + 1 rather than 1 - x.

Using Pythagoras:
%28x%2B1%29%5E2=x%5E2%2B%282x-1%29%5E2

Expand the binomials, collect terms, and solve
x%5E2%2B2x%2B1=x%5E2%2B4x%5E2-4x%2B1
4x%5E2-6x=0
x%284x-6%29=0

So x=0 or x=3%2F2. But we can exclude the x=0 root because that would mean that all three sides of the triangle were of length 0 -- a very uninteresting triangle indeed. So, the short leg is 3%2F2cm, the hypotenuse is 3%2F2%2B1=5%2F2cm and the long leg is 2%283%2F2%29-1=3-1=2cm.

Check:

sqrt%28%283%2F2%29%5E2%2B%284%2F2%29%5E2%29%29 =
sqrt%28%289%2F4%29%2B%2816%2F4%29%29 =
sqrt%2825%29%2Fsqrt%284%29=5%2F2, Check!

Hope that helps.
John

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The legs are
x, 2x+-+1
The hypotenuse is x+%2B+1
By the pythagorean teorem:
%28x+%2B+1%29%5E2+=+x%5E2+%2B+%282x+-+1%29%5E2
x%5E2+%2B+2x+%2B+1+=+x%5E2+%2B+4x%5E2+-+4x+%2B+1
2x+%2B+1+=+4x%5E2+-+4x+%2B+1
4x%5E2+-+6x+=+0
2x%282x+-+3%29+=+0
One of the factors has to be zero. Either 2x+=+0
or 2x+-+3+=+0
If 2x+=+0, then x+=+0 and then one of the legs would be zero length, so 2x+-+3+=+0
2x+=+3
x+=+3%2F2cm shortest leg
2x+-+1+=+2%2A%283%2F2%29+-+1
2x+-+1+=+2cm other leg
x+%2B+1+=+%283%2F2%29+%2B+1
x+%2B+1+=+5%2F2cm hypotenuse
The sides are 3/2, 2, 5/2
check:
%285%2F2%29%5E2+=+%283%2F2%29%5E2+%2B+2%5E2
25%2F4+=+%289%2F4%29+%2B+%2816%2F4%29
25%2F4+=+25%2F4
OK