<PRE><B>Question 14834</B>
Let x = base of triangle
x-3 = height of triangle


Area = {{{(1/2)*b*h = 20}}}
{{{(1/2)*x*(x-3) = 20 }}}


Multiply both sides by 2 to clear the fraction:
{{{2*(1/2)*x*(x-3) = 2*20}}}
{{{x^2 - 3x = 40 }}}


Quadratic equation, so set equal to zero:
{{{x^2 - 3x - 40 = 0}}}
{{{ (x-8)(x+5) = 0}}}
x=8 base
x-3= 5 height

Reject x=-5, since the base of a triangle cannot be negative.


R^2 at SCC</PRE>