What is a Conical Frustum?

Geometry Level 3

Imagine an perfect hourglass that has a base radius R 1 R_1 and a height Z Z from base to center. A solid ball with a radius of R 2 R_2 is stuck perfectly within the middle of the hourglass (it leaves no gaps). The hourglass is filled completely with sand on both sides of the hourglass. Given that

  • Z = 3 R 1 R 2 π cm Z = \dfrac {3}{R_1R_2\pi} \text{ cm} .
  • R 1 + R 2 = 9 R 1 R 2 cm R_1+R_2 = 9\sqrt {R_1R_2}\text{ cm} .
  • The volume of the sand is 120 cm 3 120\text{ cm}^3 .

Find the volume of the ball.

Note : Image drawn not up to scale.


The answer is 40.

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Anthony Ling by Anthony Ling , 24, USA

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1 solution

Anthony Ling
May 30, 2016

The volume of the ball can be found using the equation V b = V h − V s V_b = V_h - V_s , where V b V_b is the volume of the ball, V h V_h is the volume of the hourglass, and V s V_s is the volume of the sand.

Since the hourglass consists of two conical frustums (or just cones with their tops chopped off): V h = 2 V c = 2 3 π Z ( R 1 2 + R 1 R 2 + R 2 2 ) = 2 3 π Z ( ( R 1 + R 2 ) 2 − R 1 R 2 ) V_h = 2V_c = \frac {2}{3} \pi Z (R_1^2 + R_1R_2 + R_2^2) = \frac {2}{3} \pi Z ((R_1+ R_2)^2 - R_1R_2)

Plugging all the known variables in: V h = 2 3 π ( 3 R 1 R 2 π ) ( ( 9 R 1 R 2 ) 2 − R 1 R 2 ) = 2 R 1 R 2 ( 80 R 1 R 2 ) = 160 V_h = \frac {2}{3} \pi (\frac {3}{R_1R_2 \pi}) ((9 \sqrt {R_1R_2})^2 - R_1R_2) = \frac {2}{R_1R_2} (80 R_1R_2) = 160

Plugging the remaining variables, we get V b = 160 − 120 = 40 V_b = 160 - 120 = \boxed {40}

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