Question 1152767

In a right triangle ABC, 

angle {{{C = 90 }}}

{{{c=3}}}

{{{(a+b)^2=17}}}



the area of the triangle is: {{{A=(1/2)a*b}}}

using Pythagorean theorem,

{{{a^2+b^2=c^2}}}
{{{a^2+b^2=9}}}.......eq.1

{{{(a+b)^2=17}}}
{{{a^2+2ab+b^2=17}}}

{{{a^2+b^2=17-2ab}}}.....eq.2

from eq.1 and eq.2 we have

{{{9=17-2ab}}}....solve for {{{a}}}

{{{2ab=17-9}}}

{{{2ab=8}}}

{{{ab=4}}}.......eq.1a

go to

the area of the triangle is: {{{A=(1/2)a*b}}}, substitute {{{ab}}}

 {{{A=(1/2)4}}}

{{{A=2}}}