First, let's consider what information we have.
We are given a right circular cone with a radius of 10 km
and a diagonal of 22.4 km.
There are two formulae that we need: one for its volume and
one for its surface area.
The volume of a cone is [tex]V = \frac{1}{3}Bh[/tex], where B is the
area of the base and h is the height. The base of a cone is a circle, so we use
the area of a circle, [tex] \pi r^2[/tex], to get [tex]V =\frac{1}{3}\pi r^2h[/tex].
We aren’t given the height explicitly; however, using the Pythagorean
Theorem, we can find it. Treat the diagonal as the hypotenuse of a right
triangle and the radius as a leg. We have
[tex](22.4)^2-10^2=h^2\\h = 20.0[/tex].
Now, we can substitute our values into the formula for volume.
[tex]V= \frac{1}{3} \pi r^2h = \frac{1}{3} \pi (10)^2(20)= \frac{2000 \pi }{3}=2094 \ km^3[/tex]
For the surface area, we have the formula [tex] \pi rl+ \pi r^2[/tex], where l is the diagonal that we had in the beginning.
So, we simply write [tex] \pi (10)(22.4)+ \pi (10)^2=1018 \ km^2[/tex].
