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If the volume of the cube be equal to $432$, what is the volume of pyramid $ABCD$ ?

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$1)32\sqrt3\qquad\qquad2)32\sqrt[3]2\qquad\qquad3)72\qquad\qquad4)144$

To solve the problem I denoted the side of cube with $a$. and the volume of the pyramid $ABCD$ is equal to Area of $$\frac13\times S_{\triangle ABD}\times CH$$ Where $CH$ is a perpendicular segment form $C$ to the plane $ABD$.

And $\triangle ABD$ is equilateral triangle with the sides equal to $a\sqrt2$ hence $S_{\triangle ABD}=\frac{\sqrt3}4(a\sqrt2)^2=\frac{\sqrt3}2a^2$

But I don't know how to find $CH$.

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    $\begingroup$ I think it’s easier to take triangle ABC as the base. $\endgroup$
    – paulinho
    May 25, 2021 at 19:31
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    $\begingroup$ rule of thumb: area of triangle is $\dfrac1{2!}$ area of rectangle; volume of pyramid is $\dfrac1{3!}$ volume of cube $\endgroup$ May 25, 2021 at 19:34
  • $\begingroup$ @J.W.Tanner Nice! Does this pattern of $\frac1{n!}$ continues? I mean do we have ratio of $\frac1{4!}$ for something else related? $\endgroup$
    – Etemon
    May 25, 2021 at 19:38
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    $\begingroup$ @Soheil: yes; cf. this question $\endgroup$ May 25, 2021 at 19:53

1 Answer 1

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Using the formula $\text{Volume} = \dfrac{1}{3} \cdot \text{Base} \cdot \text{Height}$ is a great idea, but instead of using $\Delta ABD$ as the base and $CH$ as the height, try using $\Delta ABC$ as the base and $CD$ as the height.

Clearly, $CD = a$, and $\Delta ABC$ is a right triangle with legs $a$ and $a$, so the area of $\Delta ABC$ is $\dfrac{1}{2}a^2$

Then, the volume of the pyramid is $\text{Volume} = \dfrac{1}{3} \cdot \text{Base} \cdot \text{Height} = \dfrac{1}{3} \cdot \dfrac{1}{2}a^2 \cdot a = \dfrac{1}{6}a^3$.

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