Question

A conical cup is made from a circular piece of paper with radius 10 cm by cutting out a sector and joining the edges as shown below. Suppose θ = 9π/5. (a) Find the circumference C of the opening of the cup. (b) Find the radius r of the opening of the cup. c) Find the height h of the cup. (d) Find the volume V of the cup.

Answer 1

Part a:

The opening of the cup is the circular base of the cup which has a circumference equal to the length of the arc formed by angle θ = 9π/5 on the circular piece of paper from which the cone was made.

Thus, the circumference of the circle = Length of the arc formed by angle θ = 9π/5 at the center which is given by
$C=r\theta \\ \\ =10\times \frac{9\pi}{5} \\ \\ =18\pi\approx56.55\ cm$


Part b:
The opening of the cup is the circular base of the cup which has a circumference equal to the length of the arc formed by angle θ = 9π/5 on the circular piece of paper from which the cone was made.

Recall that the circumference of a circle is given by $C=2\pi r$ and having obtained from part a that the circumference of the circular opening is $18\pi$ cm.

Thus,
$2\pi r=18\pi \\ \\ \Rightarrow r=9\ cm$


Part c:
The height of the cup can be obtained by noticing that the radius, height and the slant height of the cup forms a right triangle with the height and the radius as the legs and the slant height as the hypothenus.

Using pythagoras theorem, the height of the cup is obtained as follows:
$h^2+r^2=l^2$
where: h is the height, r is the radius and l is the slant height.

$h^2+9^2=10^2 \\ \\ \Rightarrow h^2=100-81=19 \\ \\ \Rightarrow h= \sqrt{19} \approx4.36\ cm$


Part d
Recall that the volume of a cone is given by $V= \frac{1}{3} \pi r^2h$

Thus, the volume of the cup is given by
$V= \frac{1}{3} \pi\times9^2\times \sqrt{19} \\ \\ \approx369.7\ cm^3$