Question 1151970
What is the area of an isosceles right triangle with perimeter {{{8+4sqrt(2)}}}?

Isosceles Right Triangle:
A right triangle with the two legs (and their corresponding angles) equal. 

the perimeter of an isosceles triangle, the expression {{{P=2s + b}}} is used, where {{{s}}} represents the length of the two congruent sides and {{{b}}} represents the length of the base


 {{{ drawing( 600, 600, -10, 10, -10, 10,
circle(-5,-5,.12), circle(5,-5,.12), locate(3,-2.5,s), locate(-2.5,-2.5,s),
line(-5,-5,5,-5),locate(1.5,-4,b),line(-5,-5,0,0),line(0,0,5,-5),
 graph( 600, 600, -10, 10, -10, 10, 0)) }}}

use the Pythagorean Theorem, which states that {{{s^2 + s^2 = b^2}}}, to  express {{{b}}} in terms of {{{s}}}

{{{2s^2 = b^2}}}

{{{sqrt(2s^2) = b}}}

{{{b=s*sqrt(2)}}}

then substitute  in
{{{P=2s + b}}}

{{{8+4sqrt(2)=2s + s*sqrt(2)}}}

{{{8+4sqrt(2)=s(2+ sqrt(2))}}}

{{{s=(8+4sqrt(2))/(2+ sqrt(2))}}}

{{{s=4(2+sqrt(2))/(2+ sqrt(2))}}}

{{{s=4cross((2+sqrt(2)))/cross((2+ sqrt(2)))}}}

{{{s=4}}}

 then the area of an isosceles right triangle is:

{{{A=(1/2)s*s}}}

{{{A=s^2/2}}}

{{{A=4^2/2}}}

{{{A=16/2}}}

{{{A=8}}}