Answer :
we have that
the area of the figure is equal to the area of a right triangle plus the area of a semicircle
so
[tex]A=\frac{1}{2}\cdot b\cdot h+\frac{1}{2}\cdot\pi\cdot r^2[/tex]where
Applying Pythagorean Theorem
Find out the base of the triangle
5^2=4^2+b^2
b^2=25-16
b^2=9
b=3 in
h=4 in
r=2 in
substitute given values
[tex]A=\frac{1}{2}\cdot3\cdot4+\frac{1}{2}\cdot\pi\cdot2^2[/tex]