Depth of the Ice Cream in a Cone

Question

I received the following question in maths today and I don't know how to tackle it.

"The volume of the ice-cream is half the volume of the cone. The cone has a 3cm radius and height of 14cm. What is the depth of the ice-cream?"

I know the volume of the ice cream is $21\pi$, but I don't know how to figure out the height of the ice-cream. When I use the equation I get the following: $$21\pi = \frac{\pi r^2h}{3}$$ $$63 = r^2h$$

But I have know idea about how to ascertain the height as the radius (and of course the height) is different in the "new" cone.

Anyone know how to do it?

Answer 1

Hint Consider the bottom empty cone with so-far unknown height $h'$ and radius $r'$. We know that is has half the volume of the whole cone, i.e. $$\frac{21 \pi}{2}= \frac{\pi (r')^2 h'}{3}.$$

Also, the bottom cone has the same angle as the whole cone, which means that $\frac{r}{h}=\frac{r'}{h'}$.

Can you use this to solve for $h'$?

Answer 2

No formula required.

As the two cones are similar, the ratio of the volumes is the cube of the ratio of the sizes.

$$\frac{v_0}{v_1}=\left(\frac{h_0}{h_1}\right)^3$$

To get half the volume, you divide the total height by $\sqrt[3]2$. (The given upper radius is of no use.)