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 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 formed. In the right - triangle with an angle of $30^{\circ}$ and hypotenuse $4\sqrt{3}\text{ cm}$, using the property that in a $30 - 60-90$ triangle, if the hypotenuse is $c$, the side opposite the $30^{\circ}$ angle is $\frac{c}{2}$ and the side opposite the $60^{\circ}$ angle is $\frac{\sqrt{3}c}{2}$. Here, the height $h$ of the pyramid (opposite the $30^{\circ}$ angle) is $4\text{ cm}$.

Step3: Calculate the volume

Substitute $B = 12\sqrt{3}\text{ cm}^2$ and $h = 4\text{ cm}$ into the volume formula $V=\frac{1}{3}Bh$. So $V=\frac{1}{3}\times12\sqrt{3}\times4=16\sqrt{3}\text{ cm}^3$.

Answer:

$16\sqrt{3}\text{ cm}^3$