Question 1150800
<pre>{{{drawing(28000/83,400,-5,135,-5,161,locate(14,50,D),
locate(0,0,A), locate(130,0,B), locate(67,160,C),
line(0,0,130,0),line(130,0,65,156),line(0,0,65,156),green(line(130,0,19.23076923,46.15394615)), locate(60,0,130), locate(60,36,120)


 )}}} 

The green line BD is the altitude drawn from vertex of base angle B, so BD is
perpendicular to AC. So ADB is a right triangle, so by the Pythagorean theorem,

{{{AD^2+BD^2=AB^2}}}
{{{AD^2+120^2=130^2}}}
{{{AD^2=130^2-120^2}}}
{{{AD^2=16900-14400}}}
{{{AD^2=2500}}}
{{{AD=sqrt(2500)}}}
{{{AD=50}}}

CDB is also a right triangle, so also by the Pythagorean theorem,

{{{BD^2+CD^2=BC^2}}}
{{{120^2+CD^2=BC^2}}}

But since BC = AD + CD
          BC = 50 + CD

{{{120^2+CD^2=(50+CD)^2}}}
{{{120^2+CD^2=50^2+100CD+CD^2}}}
{{{120^2=50^2+100CD}}}
{{{14400-2500=100CD}}}
{{{11900=100CD}}}
{{{119=CD}}}

{{{AC = AD + CD}}}
{{{AC = 50 + 119}}}
{{{AC = 169}}}

{{{Area=expr(1/2)base*altitude}}}
{{{Area=expr(1/2)AC*BD}}}
{{{Area=expr(1/2)169*120}}}
{{{Area=matrix(1,2,10140,m^2)}}}


Edwin</pre>