Question 850285
Let {{{ x }}} = one of the dimensions of the rectangle
{{{ 60/x }}} = the other dimension
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The center of the circle that this rectangle is
inscribed in is located where the diagonals cross
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The length of a diagonal is:
{{{ d = sqrt( ( 60/x )^2 + x^2 ) }}}
{{{ d = sqrt( 3600/x^2 + x^2 ) }}}
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This diagonal is the same as the diameter of the
circle, which is
{{{ 2r = 2*6.5 }}} 
{{{ 2r = 13 }}}
Now I can say:
{{{ 13 = sqrt( 3600/x^2 + x^2 ) }}}
Square both sides
{{{ 169 = 3600/x^2 + x^2 }}}
Let {{{ x^2 = z }}}
{{{ 169 = 3600/z + z }}}
Multiply both sides by {{{ z }}}
{{{ 169z = 3600 + z^2 }}}
{{{ z^2 - 169z + 3600 }}}
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{{{ z = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 1 }}}
{{{ b = -169 }}}
{{{ c = 3600 }}}
{{{ z = ( 169 +- sqrt( (-169)^2 - 4*1*3600 )) / (2*1) }}}
{{{ z = ( 169 +- sqrt( 28561 - 14400 )) / 2 }}}
{{{ z = ( 169 +- sqrt( 14161 )) / 2 }}}
{{{ z = ( 169 - 119) / 2 }}}
{{{ z =  50/ 2 }}}
{{{ z = 25 }}}
and
{{{ z = x^2 }}}
{{{ x = sqrt(25) }}}
{{{ x = 5 }}}
{{{ 60/x = 60/5 }}}
{{{ 60/x = 12 }}}
The dimension are:
 5 x 12 in2
check:
{{{ 5*12 = 60 }}}
OK
Note that is you go back and pick the
positive {{{ sqrt( 14161 ) }}}, then you
get the same answers