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The longer leg of a right triangle is 1 ft longer than the shorter leg. The hypotenuse is 1 ft shorter than twice the shorter leg. Find the length of the shorter leg of the triangle.
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one leg x, the other is x + 1 and the hypotenuse is 2(x-1) now pythathatoas
\[x^2+(x+1)^2=(2(x-1))^2\]
\[x^2+x^2+2x+1=4(x^2-2x+1)\] \[2x^2+2x+1=4x^2-8x+4\] \[2x^2-10x+3=0\]
oh damn i did it wrong! says hypotenuse is 1 ft shorter than twice the shorter leg so it should be 2x-1 i start again
\[x^2+(x+1)^2=(2x-1)^2 \]\[2x^2+2x+1=4x^2-4x+1\] \[2x^2-6x=0\] \[2x(x-3)=0\] so x = 0 which is silly, or x = 3 which gives a nice 3 - 4 - 5 right triangle which we could have guessed from the beginning
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I get it thank you for your help
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