Inverted Cone

Geometry Level 1

The top of a cone of radius 10 is inverted so that the very top of the cone touches the base. The height of the resulting frustum is 3. The frustum is pictured below.

What is the volume of the space in between the lateral surface of the frustum and the lateral surface of the cone inserted into it?

NOTE: for a right circular cone, V = π r 2 h 3 V=\pi r^2 \frac{h}{3}

200π 150π 50π 100π

Caleb Koch by Caleb Koch , 23, USA

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1 solution

Consider the figure above. By ratio and proportion, we have

r 3 = 10 6 \dfrac{r}{3}=\dfrac{10}{6} ⟹ \implies r = 5 r=5

The volume of the small cone is : 1 3 π ( 5 2 ) ( 3 ) = 25 π \dfrac{1}{3}\pi (5^2)(3)=25\pi .

The volume of the big cone is: 1 3 π ( 1 0 2 ) ( 6 ) = 200 π \dfrac{1}{3}\pi (10^2)(6)=200 \pi

The volume of the frustum is volume of the big cone minus volume of the small cone: 200 π − 25 π = 175 π 200 \pi - 25\pi=175 \pi

Now, we need to subtract again the volume of the small cone, so the desired volume is

175 π − 25 π = 175 \pi - 25 \pi= 150 π \boxed{150 \pi}

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