Question 318709
1.{{{A=(1/2)L1*L2=60}}}
2.{{{H=L2+2}}}
From the Pythagorean theorem,
3.{{{L1^2+L2^2=H^2}}}
From eq. 1,
{{{L1*L2=120}}}
{{{L1=120/L2}}}
Substitute into eq. 3,
{{{(120/L2)^2+L2^2=(L2+2)^2}}}
{{{(120/L2)^2+L2^2=L2^2+4L2+4}}}
{{{(14400/L2^2)=4L2+4}}}
{{{14400=4L2^3+4L^2}}}
{{{L2^3+L2^2-3600=0}}}
You can factor this equation,
{{{(L2-15)(L2^2+16L2+240)=0}}}
One real solution:
{{{L2-15=0}}}
{{{highlight( L2=15)}}}
.
.
{{{L1=120/15}}}
{{{highlight( L1=8)}}}
.
.
{{{H=L2+2}}}
{{{highlight( H=17 )}}}