Question 738771
If it's isosceles, then both legs are equal
Let {{{ x }}} = length of one of the legs
{{{ x^2 + x^2 = ( x + 1 )^2 }}}
{{{ 2x^2 = x^2 + 2x + 1 }}}
{{{ x^2 - 2x - 1 = 0 }}}
Use quadratic formula
{{{ x = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 1 }}}
{{{ b = -2 }}}
{{{ c = -1 }}}
{{{ x = (-(-2) +- sqrt( (-2)^2 - 4*1*(-1) )) / (2*1) }}}
{{{ x = ( 2 +- sqrt( 4 + 4 )) / 2 }}}
{{{ x = ( 2 +- sqrt( 8 )) / 2 }}}
{{{ x = ( 2 + 2*sqrt(2) ) / 2 }}}
{{{ x = 1 + sqrt(2) }}} answer
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check:
{{{ 2x^2 = ( x + 1 )^2 }}}
{{{ 2*( 1 + sqrt(2) )^2 = ( 2 + sqrt(2) )^2 }}}
{{{ 2*( 1 + 2*sqrt(2) + 2 ) = 4 + 4*sqrt(2) + 2 }}}
{{{ 6 + 4*sqrt(2) = 6 + 4*sqrt(2) }}}
OK