document.write( "Question 119171This question is from textbook
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\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "i have:
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\n" ); document.write( "1-x
\n" ); document.write( "2x-1
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Algebra.Com's Answer #87278 by solver91311(24713)\"\" \"About 
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.\r
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\n" ); document.write( "\n" ); document.write( "Using Pythagoras:
\n" ); document.write( "\"%28x%2B1%29%5E2=x%5E2%2B%282x-1%29%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "Expand the binomials, collect terms, and solve
\n" ); document.write( "\"x%5E2%2B2x%2B1=x%5E2%2B4x%5E2-4x%2B1\"
\n" ); document.write( "\"4x%5E2-6x=0\"
\n" ); document.write( "\"x%284x-6%29=0\"\r
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\n" ); document.write( "\n" ); document.write( "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%2F2\"cm, the hypotenuse is \"3%2F2%2B1=5%2F2\"cm and the long leg is \"2%283%2F2%29-1=3-1=2\"cm.\r
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\n" ); document.write( "\n" ); document.write( "Check:\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%28%283%2F2%29%5E2%2B%284%2F2%29%5E2%29%29\" =
\n" ); document.write( "\"sqrt%28%289%2F4%29%2B%2816%2F4%29%29\" =
\n" ); document.write( "\"sqrt%2825%29%2Fsqrt%284%29=5%2F2\", Check!\r
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\n" ); document.write( "\n" ); document.write( "Hope that helps.
\n" ); document.write( "John
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