A geometry problem by A Former Brilliant Member

Geometry Level 2

A right truncated prism has a horizontal triangular base A B C ABC . A B AB is 10 10 cm, B C BC is 12 12 cm and C A CA is 8 8 cm. The vertical edges through A , B , A,B, and C C are 20 20 cm, 12 12 cm and 18 18 cm long respectively. Determine the volume of the solid in cubic centimeters rounded to the nearest integer.

Note: The vertical edges are all perpendicular to the base.


The answer is 661.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

The volume of a right truncated prism is equal to the area of the right section multiplied by the average lateral edge.

V = K m \boxed{V=Km} where: K K =area of the right section and m m =average lateral edge

Solving for K:

s = 10 + 12 + 8 2 = 15 s=\frac{10+12+8}{2}=15

K = A T = 15 ( 15 − 10 ) ( 15 − 12 ) ( 15 − 8 ) = 1575 K=A_T = \sqrt{15(15-10)(15-12)(15-8)}=\sqrt{1575}

Solving for m:

m = 20 + 12 + 18 2 = 50 3 m=\frac{20+12+18}{2}=\frac{50}{3}

Solving for V:

V = 1575 ( 50 3 ) = 661 c m 3 V=\sqrt{1575}(\frac{50}{3})=661~cm^3

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