Question 1108259
Let x = width
x-12 = base
c= hypotenuse 
Perimeter = x + x-12 + c = 144
2x+c -12 = 144
2x+c= 156
c=156-2x


Since it's a right triangle, {{{a^2 + b^2 = c^2}}}
{{{x^2 + (x-12)^2 = (156-2x)^2}}}
{{{ x^2 + x^2-24x + 144 = 156^2 - 624x + 4x^2 }}}
{{{2x^2 -24x +144= 24336 -624x+4x^2}}}


{{{0= 2x^2-600x + 24192}}}
{{{0= x^2 -300x + 12096}}}


Rewrite this to solve by completing the square 

{{{x^2 -300 + ____=-12096 + ______}}}
{{{x^2 - 300 + 22500 = -12096 + 22500}}}


{{{ (x-150) ^2 = 10404}}}


Take square root of both sides:
{{{x-150=0 +-102}}}
{{{x=150 +-102}}}

{{{x=252}}}  or  {{{x=48}}}


The first answer is an extraneous root and must be rejected since the perimeter exceeds 144.  The second answer gives legs of a right triangle to be 48 and 36, so the area is {{{A= 1/2 *48*36 = 864 cm^2}}}   


Dr. Robert J. Rapalje, Retired
Email:  rapaljer@mathinlivingcolor.com


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