Question 1068027
h-hypotenuse
b-base
a-height
A-area
.
.

{{{h=4b}}}
You also know,
{{{a^2+b^2=h^2}}}
{{{a^2+b^2=(4b)^2}}}
{{{a^2=16b^2-b^2}}}
{{{a^2=15b^2}}}
{{{a=sqrt(15)b}}}
So then the area,
{{{A=(1/2)ba}}}
{{{A=(1/2)b*sqrt(15)b}}}
{{{A=(sqrt(15)/2)b^2}}}