Question 550143
think of the 5-5-6 triangle with the 6 side horizontal
The area is {{{ (1/2)*6*h[1] }} where {{{h[1]}}} is the height
{{{ h[1] = 4 }}} since the height is a leg of
a 3-4-5 right triangle
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The 5-5-x triangle with x side horizontal
The area is {{{ (1/2)*x*h[2] }}}
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given:
{{{ (1/2)*6*h[1] = (1/2)*x*h[2] }}}
{{{ 6h[1] = x*h[2] }}}
{{{ 6*4 = x*h[2] }}}
{{{ x = 24/h[2] }}}
{{{ x^2 = 576/(h[2])^2 }}}
{{{ (h[2])^2 }}} in terms of {{{x^2}}}  is
{{{ 5^2 = (x/2)^2 + (h[2])^2 }}}
{{{ (h[2])^2 = 25 - x^2/4 }}}
and
{{{ x^2*(25 - x^2/4) = 576 }}}
{{{ 25x^2 - x^4/4 = 576 }}}
{{{ x^4 - 100x^2 + 2304 = 0 }}}
Let {{{x^2 = z }}}
{{{ z^2 - 100z + 2304 = 0 }}}
Complete the square
{{{ z^2 - 100z + 2500 = 2500 - 2304 }}}
{{{ ( z - 50 )^2 = 196 }}}
{{{ z - 50 = 14 }}}
{{{ z = 64 }}}
{{{ x^2 = 64 }}}
{{{ x = 8 }}}
x is 8 cm
check answer:
{{{ x = 24/h[2] }}}
{{{ 8 = 24/h[2] }}}
{{{ h[2] = 3 }}}
and
{{{ (h[2])^2 = 25 - x^2/4 }}}
{{{ 3^2 = 25 - 8^2/4 }}}
{{{ 9 = 25 - 64/4 }}}
{{{ 9 = 25 - 16 }}}
{{{ 9 = 9 }}}
OK
and
{{{ (1/2)*6*h[1] = (1/2)*x*h[2] }}}
{{{ 6*4 = 8*3 }}}
{{{ 24 = 24 }}}
OK