Question

a solid oblique pyramid has an equilateral triangle as a base with an edge length of 4\sqrt{3} cm and an area of 12\sqrt{3} cm². what is the volume of the pyramid? 12\sqrt{3} cm³ 16\sqrt{3} cm³ 24\sqrt{3} cm³ 32\sqrt{3} cm³

Explanation:

Step1: Recall volume formula for pyramid

The volume formula for a pyramid is $V=\frac{1}{3}Bh$, where $B$ is the area of the base and $h$ is the height.

Step2: Identify base - area and find height

We are given that the area of the base $B = 12\sqrt{3}\text{ cm}^2$. To find the height, consider the right - triangle $ABC$. In right - triangle $ABC$, $\angle A=30^{\circ}$ and the hypotenuse $AB = 4\sqrt{3}\text{ cm}$. Using the property of a 30 - 60 - 90 triangle ($\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$), if $\theta = 30^{\circ}$, then the height $h$ (opposite to the $30^{\circ}$ angle) is $h = 2\sqrt{3}\text{ cm}$.

Step3: Calculate volume

Substitute $B = 12\sqrt{3}\text{ cm}^2$ and $h = 2\sqrt{3}\text{ cm}$ into the volume formula $V=\frac{1}{3}Bh$.
\[

$$\begin{align*} V&=\frac{1}{3}\times12\sqrt{3}\times2\sqrt{3}\\ &=\frac{1}{3}\times12\sqrt{3}\times2\sqrt{3}\\ &=\frac{1}{3}\times12\times2\times3\\ & = 24\sqrt{3}\text{ cm}^3 \end{align*}$$

\]

Answer:

$24\sqrt{3}\text{ cm}^3$