# Volume of a frustum given the bottom radius and the top cone height.

A cone with base radius 12 cm is sliced parallel to its base, as shown, to remove a smaller cone of height 15 cm. If the height of the smaller cone is three-fourths that of the original cone, what is the volume of the remaining frustum?

I set the frustum's top radius as x and its height as y. Using the cone volume formula, I have $$\frac{144\pi\cdot(y+15)}{4}=\frac{15x^2\pi}{3}$$$$\implies 36 y+540=5x^2$$

I am stuck here. How should I continue?

For the height of the bigger cone: We have $$h=15 = \frac{3}{4}H \implies H = 20\text {cm}$$

$$\tan z = \frac{r}{h} = \frac{R}{H}$$

We have $$R = 12, H = 20 , h = 15 \text{ (all in cm) } \implies r = \frac{R}{H}h = \cdots$$

Now the height of the frustrum is $$H-h = \cdots$$

• Oh gawd. I read the problem incorrectly and couldn't see that. Thanks. Aug 16, 2019 at 23:55
• You're welcome! Aug 16, 2019 at 23:55

Hint:

The smaller cone is homothetic of the larger cone in a homothety with centre the vertex of the cones and ratio $$3/4$$. So the height $$h$$ of the smaller cone is $$h=3/4$$ of the height $$H$$ of the larger cone, its base area is $$b=9/16B$$ and its volume is $$v=27/64V$$. So the remaining frustum has volume $$\mathcal V=V-v =\frac{37}{64}V.$$

Can you calculate the volume of the given cone?