Question 1138788
The height bisects the angle between 10m sides
Call these angles both {{{ theta }}}
The height is {{{ 10*cos( theta ) }}}
one-half of the base is {{{ 10*sin( theta ) }}}
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The area is:
{{{ A = (1/2)*( 10*sin(theta) )*10*cos(theta) ) }}}
{{{ A = 50*sin(theta)*cos(theta) }}}
Use trig identity
{{{ A = 50*sin( 2theta )}}}
The rate of change is the cos, so
{{{ A[prime] = 50*cos( 2theta ) }}}
Max is where {{{ A[prime] = 0 }}}
{{{ 50*cos( 2theta ) = 0 }}}
{{{ 2theta = pi/2 }}}
{{{ A[max] = 50*sin(pi/2) }}}
{{{ A[max] = 50 }}} m2
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The angles are {{{ pi/2 }}}, {{{ pi/4 }}}, {{{ pi/4 }}}
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Get a 2nd opinion if needed. 
Check my math, too