Question 1057613
{{{H=2A}}}
{{{B=sqrt(12)}}}
{{{A^2+B^2=H^2}}}
Substituting,
{{{A^2+12=(2A)^2}}}
{{{A^2+12=4A^2}}}
{{{3A^2=12}}}
{{{A^2=4}}}
{{{A=2}}}{{{ft}}}
So,
{{{H=2(2)}}}
{{{H=4}}}{{{ft}}}