# The volume of cube is given how to find the volume of pyramid $ABCD$?

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

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

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 $$CH$$.

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

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 $$\text{Volume} = \dfrac{1}{3} \cdot \text{Base} \cdot \text{Height} = \dfrac{1}{3} \cdot \dfrac{1}{2}a^2 \cdot a = \dfrac{1}{6}a^3$$.