Question 952730

in a right triangle:{{{c^2=a^2+b^2}}} 
 
if the sides of a right triangle are {{{x}}}, {{{x+31}}}, and {{{x+32}}} units long, then we have

{{{(x+32)^2=x^2+(x+31)^2}}} 

{{{x^2+64x+32^2=x^2+x^2+62x+31^2}}} 

{{{x^2+64x+1024=2x^2+62x+961}}} 

{{{0=2x^2-x^2+62x-64x+961-1024}}}

{{{x^2-2x-63=0}}}

{{{x^2+7x-9x-63=0}}}

{{{(x^2+7x)-(9x+63)=0}}}

{{{x(x+7)-9(x+7)=0}}}

{{{(x+7)(x-9)=0}}}

solutions:

if {{{(x+7)=0}}}=>{{{x=-7}}}...disregard negative solution, side the length must be positive

if {{{ (x-9)=0}}}=>{{{highlight(x=9)}}}-> your answer

so, now we can find the length of all three sides:

{{{highlight(x=9)}}}

{{{highlight(x+31=40)}}}

{{{highlight(x+32=41)}}}


check if they are the sides of a right-angle triangle:

{{{(x+32)^2=x^2+(x+31)^2}}}

{{{41^2=9^2+40^2}}}

{{{1681=81+1600}}}

{{{1681=1681}}}...which verifies our answer