Question 1040514
Let {{{ A }}} = the area of the right triangle in square inches
Let {{{ b }}} = the base of the right triangle in inches
Let {{{ c }}} = the height of the right triangle in inches
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{{{ A = (1/2)*b*c }}}
{{{ 24 = (1/2)*b*c }}}
{{{ b*c = 48 }}}
{{{ c = 48/b }}}
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Use pythagorean theorem
{{{ 12^2 = b^2 + c^2
{{{ 144 = b^2 + c^2
By substitution:
{{{ b^2 + ( 48/b )^2 = 144 }}}
{{{ b^4 + 2304 = 144b^2 }}}
{{{ 144b^2 - b^4 = 2304 }}}
Let {{{ z = b^2 }}}
{{{ -z^2 + 144z - 2304 = 0 }}}
Use quadratic formula
{{{ z = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = -1 }}}
{{{ b = 144 }}}
{{{ c = -2304 }}}
{{{ z = ( -144 +- sqrt( 144^2-4*(-1)*(-2304) ))/(2*(-1)) }}}
{{{ z = ( -144 +- sqrt( 20736 - 9216 ))/((-2) ) }}}
{{{ z = ( -144 +- sqrt( 11520 ) / (-2) ) }}}
{{{ z = ( -144 + 107.3313 ) / (-2) }}}
{{{ z = 36.6687/2 }}}
{{{ z = 18.3344 }}}
and, also:
{{{ z = ( -144 - 107.3313 ) / (-2) }}}
{{{ z = 251.3313/2 }}}
{{{ z = 125.6657 }}}
Since {{{ z = b^2 }}}
{{{ b = sqrt(z) }}}
{{{ b = sqrt( 18.3344 ) }}}
{{{ b = 4.2819
and
{{{ c = 48/b }}}
{{{ c = 48/4.2819 }}}
{{{ c = 11.21 }}}
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check:
{{{ 24 = (1/2)*b*c }}}
{{{ 48 = b*c }}}
{{{ 48 = 4.2819*11.21 }}}
{{{ 48 = 48.0001 }}}
close enough
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{{{ 144 = b^2 + c^2
{{{ 144 = 4.2819^2 + 11.21^2
{{{ 144 = 18.3347 + 125.6641 }}}
{{{ 144 = 143.9988 }}}
close enough
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The triangle is:
12 x 4.3 x 11.2
Hope I got it!