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# Solid right prism

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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

CPhill  Sep 16, 2017

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