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