1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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

enter image description here

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

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 $H-h = \cdots$

$\endgroup$
  • $\begingroup$ Oh gawd. I read the problem incorrectly and couldn't see that. Thanks. $\endgroup$ – Max0815 Aug 16 at 23:55
  • $\begingroup$ You're welcome! $\endgroup$ – Ak19 Aug 16 at 23:55
1
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.