Question 1151970
What is the area of an isosceles right triangle with perimeter 8 + 4√2?
<pre>Let each congruent side be s
Then with this being an isosceles right triangle, hypotenuse = {{{s * sqrt(2)}}}
We then get: {{{matrix(1,3, s + s + s * sqrt(2), "=", 8 + 4sqrt(2))}}}
{{{matrix(1,3, 2s + s * sqrt(2), "=", 8 + 4sqrt(2))}}}
Equating the 1st terms (2s = 8), or the 2nd terms ({{{matrix(1,3, s * sqrt(2), "=", 4sqrt(2))}}}, s, or one of the equal sides = {{{matrix(1,2, 4, units)}}}
With each congruent side being 4, {{{highlight_green(matrix(1,7, Area, "=", (1/2)(4)(4), "=", 8, sq, units))}}}