Question 316473
{{{A+B+26=60}}}
1.{{{A+B=34}}}
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{{{A^2+B^2=26^2}}}
2.{{{A^2+B^2=676}}}
From eq. 1,
{{{A=34-B}}}
{{{A^2=1156-68B+B^2}}}
Substitute into eq. 2,
{{{(1156-68B+B^2)+B^2=676}}}
{{{2B^2-68B+480=0}}}
{{{B^2-34B+240=0}}}
{{{(B-24)(B-10)=0}}}
Two solutions:
{{{B-24=0}}}
{{{B=24}}} 
Then
{{{A+24=34}}}
{{{A=10}}}
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{{{B-10=0}}}
{{{B=10}}}
Then
{{{A+10=34}}}
{{{A=24}}}
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Area={{{(1/2)bh=(1/2)(24)(10)=120}}}