I need help on these two problems.

Answer:
Part 1) [tex]A=16\pi\ km^{2}[/tex]
Part 2) [tex]A=196\ mi^{2}[/tex]
Step-by-step explanation:
Part 1) Find the area of the circle
we know that
The area of the circle is equal to
[tex]A=\pi r^{2}[/tex]
where
r is the radius of the circle
we have
[tex]r=4\ km[/tex]
substitute
[tex]A=\pi (4)^{2}[/tex]
[tex]A=16\pi\ km^{2}[/tex]
Part 2) Find the area of triangle
The area of triangle is equal to
[tex]A=\frac{1}{2}(b)(h)[/tex]
where
b is the base and h is the height of triangle
we have
[tex]b=20\ mi[/tex]
[tex]h=19.6\ mi[/tex]
substitute
[tex]A=\frac{1}{2}(20)(19.6)[/tex]
[tex]A=196\ mi^{2}[/tex]