Question 1190831
<pre>
Let b=base
Let h=height
Let d=diagonal

Given:
d = b+6
h = (b/3)+10

By Pythagorean theorem:
{{{d^2 = b^2 + h^2 }}}

Substitute the givens to get one eqn w/one unknown:
{{{ (b+6)^2 = b^2 + ((b/3)+10)^2 }}}

Expand (and cancel one {{{b^2}}} from each side):
{{{ cross(b^2) + 12b+36 = cross(b^2) + b^2/9 + 20b/3 + 100 }}}

Multiply both sides  by 9:
{{{ 108b+324 = b^2+60b+900 }}}

Bring it all over to one side:
{{{ b^2 -48b +576 = 0 }}}

{{{ (b-24)^2 = 0 }}}  --->  b = 24  --> h = 18,  d = 30


Check:
{{{d^2}}} = {{{30^2 }}} = 900
{{{ b^2 + h^2 = 24^2 + 18^2 = 576 + 324 = 900 }}}  (ok!)