Question 935012

hypotenuse is {{{y+2}}}

one leg is {{{y}}}, the other leg is {{{4}}}

use Pythagorean theorem to find {{{y}}}

{{{(y+2)^2=y^2+4^2}}}

{{{y^2+4y+4=y^2+16}}}

{{{cross(y^2)+4y+4=cross(y^2)+16}}}

{{{4y+4=16}}}

{{{4y=16-4}}}

{{{4y=12}}}

{{{highlight(y=3)}}}=> your answer


so, hypotenuse is {{{y+2=5}}}, one leg is {{{y=3}}}, the other leg is {{{4}}}