+168 A solid right prism $ABCDEF$ has a height of $16$ and equilateral triangles bases with side length $12,$ as shown. $ABCDEF$ is sliced with a straight cut through points $M,$ $N,$ $P,$ and $Q$ on edges $DE,$ $DF,$ $CB,$ and $CA,$ respectively. If $DM=4,$ $DN=2,$ and $CQ=8,$ determine the volume of the solid $QPCDMN.$
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I haven't seen a problem like this before, so...I'm "flying blind"....but....I believe that we will have a pryramidal frustum here......
Triangles QCP and MDN are similar.....so...
DN /DM = PC / CQ
2 / 4 = PC / 8 → PC = 4
And the volume of a pyramidal frustum is :
V = (1/3)h ( B1 + B2 + √ [ B1 * B2] 0
Here h is the frusum height, and B1 and B 2 are the areas of the bases
So the height = 16......the area of the top base , B1, = (1/2)PC * CQ sin (60°) = (1/2) (4)(8) √3/2 =
8√3 ....and the area of the bottom base, B2 = = (1/2)(DN) (DM) sin (60°) = (1/2)(2)(4) √3 / 2 =
2√3
So....the volume is
(1/3) (16) ( 8√3 + 2√3 + √ [ 8√3 * 2√3] ) =
(16/3) ( 10√3 + √48 ) =
(16/3) ( 10√3 + 4√3) =
(16/3) (14√3) = 224√3 / 3 = 224 / √3 units^2
