Question 99

# A roly-poly toy is made by joining a hemispherical bottom with a conical top. The diameter and the volume of the hemisphere are the same as the diameter of the base and volume of the cone, respectively. The toy is vertically cur at half of its height. If the diameter of the base sphere was 4 inch, what would the volume (in $$inch^{3}$$) of the upper half of the toy be?

Solution

Volume of the semi-sphere = $$\frac{1}{2}\times\ \frac{4}{3}\pi\ r^3$$
Diameter of the Semi-sphere = 2r

Volume of the Cone = $$\frac{1}{3}\times\ \pi\ r^2h$$
Diameter of the base of the cone = 2r

We are given that the diameter of the sphere is 4 inch.
2r = 4
giving us, r = 2 inches

We are also given that the volume of the cone and volume of the semi-sphere are equal.
using r =2 gives us,
$$\frac{1}{2}\times\ \frac{4}{3}\pi\ r^3$$ = $$\frac{1}{3}\times\ \pi\ r^2h$$
$$\frac{1}{2}\times\ \frac{4}{3}\pi\ 8$$ = $$\frac{1}{3}\times\ \pi\ 4h$$
this gives us h=4

Now we are to find the volume of the upper half of the toy.
The total height of the toy is 2 inches of the semi-sphere + 4 inches of the cone = 6 inches
So the upper half of this toy would be the upper 3 inches of the cone.

We will have to find the radius of base of this section of cone.
Using the tan of the semi-angle of the cone we get, $$\tan\ \theta\ =\frac{perpendicular}{base}=\frac{2}{4}=\frac{required\ base}{3}$$
This gives the required base radius as $$\frac{3}{2}$$ inches.

Finding the volume of this section of the cone, $$\frac{1}{3}\pi\ \left(\frac{3}{2}\right)^23\ =\ \frac{9\pi}{4}$$
Hence, Option D is the correct answer.