Question 1177212

let’s  hypotenuse be {{{c}}}, the shorter leg {{{b}}}, and  the other leg {{{a}}}

given:

The hypotenuse of a right triangle is three less than twice the shorter leg.

{{{c=2b-3}}}

The length of the other leg is three more than the shorter leg.

{{{a=b+3}}}

use Pythagorean theorem:

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

{{{(2b-3)^2=(b+3)^2+b^2}}}.......solve for {{{b}}}
{{{4b^2-12b+9=b^2+6b+9+b^2}}}
{{{4b^2-12b+9=2b^2+6b+9}}}
{{{4b^2-12b+9-2b^2-6b-9=0}}}
{{{2b^2-18b=0}}}
{{{2b(b-9)=0}}}

=> {{{highlight(b=9)}}}=> the length of the shorter leg


go to

{{{a=b+3}}}.......substitute {{{b}}}
{{{a=9+3}}}
{{{a=12}}}

{{{c=2b-3}}}
{{{c=2*9-3}}}
{{{c=15}}}