Question 661067
let the hypotenuse of a right triangle be {{{c}}}

and one leg {{{a}}} which is {{{9ft}}}

and the other leg {{{b}}} is {{{1 ft}}} less than the hypotenuse {{{c}}}.

so, we have:

{{{highlight(a=9ft)}}}

{{{c=b+1ft}}}...=>{{{b=c-1ft}}}

use Pythagoras theorem:

{{{c^2=a^2+b^2}}}...substitute {{{a}}} and {{{b}}} with given values

{{{c^2=(9ft)^2+(c-1ft)^2}}}

{{{c^2=81ft^2+c^2-2cft+1ft^2}}}

{{{cross(c^2)=81ft^2+cross(c^2)-2cft+1ft^2}}}

{{{0=81ft^2-2cft+1ft^2}}}

{{{2cft=81ft^2+1ft^2}}}

{{{cross(2)c*cross(ft)=cross(82)41ft^cross(2)1}}}

{{{highlight(c=41ft)}}}

now find {{{b}}}

{{{b=c-1ft}}}

{{{b=41ft-1ft}}}

{{{highlight(b=40ft)}}}


check:

{{{c^2=a^2+b^2}}}.....when {{{highlight(a=9ft)}}},{{{b=highlight(40ft)}}},{{{highlight(c=41ft)}}}


{{{(41ft)^2=(9ft)^2+(40ft)^2}}}

{{{1681ft^2=81ft^2+1600ft^2}}}

{{{1681ft^2=1681ft^2}}}