Question 1013056
You can try letting AB be one of the legs.  Base OR height is AB, and the other leg is whatever length you need for the area to be specified, 10 sq cm.


Let k be the other leg.
{{{(1/2)(AB)k=10}}}
{{{(k/2)AB=10}}}
{{{k=20/AB}}}

{{{k=20/sqrt((3-3)^2+(8-4)^2)}}}


{{{k=20/4}}}


{{{k=5}}}



AB found to be 4, and k found to be 5.
These are the triangle's dimensions, a RIGHT TRIANGLE.
Hypotenuse:  {{{sqrt((4)^2+(5)^2)}}}
{{{highlight(sqrt(41))}}}--------the hypotenuse