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? $\bigcirc\\ 12\sqrt{3}$ cm³ $\bigcirc\\ 16\sqrt{3}$ cm³ $\bigcirc\\ 24\sqrt{3}$ cm³ $\bigcirc\\ 32\sqrt{3}$ cm³

Explanation:

Step1: Find pyramid height

For the right triangle with angle $30^\circ$, opposite side (height $h$) is $\sin(30^\circ) \times$ hypotenuse $AC$.
$\sin(30^\circ) = \frac{1}{2}$, so $h = \frac{1}{2} \times 4\sqrt{3} = 2\sqrt{3}$ cm

Step2: Apply volume formula

Volume of pyramid: $\frac{1}{3} \times \text{base area} \times \text{height}$
Substitute values: $\frac{1}{3} \times 12\sqrt{3} \times 2\sqrt{3}$

Step3: Calculate final volume

First compute $\sqrt{3} \times \sqrt{3} = 3$, then $\frac{1}{3} \times 12\sqrt{3} \times 2\sqrt{3} = \frac{1}{3} \times 12 \times 2 \times 3 = 24$? No, correct calculation:
$\frac{1}{3} \times 12\sqrt{3} \times 2\sqrt{3} = \frac{1}{3} \times 12 \times 2 \times (\sqrt{3} \times \sqrt{3}) = \frac{1}{3} \times 24 \times 3 = 24$? Wait, no, recheck:
Wait, $\frac{1}{3} \times 12\sqrt{3} \times 2\sqrt{3} = \frac{1}{3} \times 12 \times 2 \times 3 = 24$? No, $12\sqrt{3} \times 2\sqrt{3} = 24 \times 3 = 72$, then $\frac{72}{3}=24$? No, wait $\sqrt{3} \times \sqrt{3}=3$, so $12 \times 2 \times 3 =72$, $\frac{72}{3}=24$? But wait, no, the options have $24\sqrt{3}$? Wait no, mistake: base area is $12\sqrt{3}$, height is $2\sqrt{3}$.
$\frac{1}{3} \times 12\sqrt{3} \times 2\sqrt{3} = \frac{1}{3} \times 12 \times 2 \times (\sqrt{3} \times \sqrt{3}) = \frac{1}{3} \times 24 \times 3 = 24$? No, wait $\sqrt{3} \times \sqrt{3}=3$, so $12\sqrt{3} \times 2\sqrt{3}=24 \times 3=72$, $\frac{72}{3}=24$. But wait, the options have $24\sqrt{3}$? Wait no, I misread the base area: base area is $12\sqrt{3}$ cm², height is $2\sqrt{3}$ cm.
Wait, $\frac{1}{3} \times 12\sqrt{3} \times 2\sqrt{3} = \frac{1}{3} \times 12 \times 2 \times (\sqrt{3} \times \sqrt{3}) = \frac{1}{3} \times 24 \times 3 = 24$ cm³? No, that can't be. Wait no, the height: $AC$ is $4\sqrt{3}$, angle at A is 30 degrees, so height $h = AC \times \sin(30^\circ) = 4\sqrt{3} \times \frac{1}{2} = 2\sqrt{3}$ cm. Correct.
Base area is $12\sqrt{3}$ cm². So volume is $\frac{1}{3} \times 12\sqrt{3} \times 2\sqrt{3} = \frac{1}{3} \times 12 \times 2 \times 3 = 24$ cm³? But the options have $24\sqrt{3}$. Wait no, I made a mistake: base area is $12\sqrt{3}$, height is $4$? No, wait the triangle: adjacent side to 30 degrees is $4\sqrt{3}$, so opposite side (height) is $\tan(30^\circ) \times$ adjacent? No! Wait, the diagram shows that AC is the horizontal side, BC is perpendicular to AC, so angle at A is 30 degrees, so $\tan(30^\circ) = \frac{BC}{AC}$, so $BC = AC \times \tan(30^\circ) = 4\sqrt{3} \times \frac{1}{\sqrt{3}} = 4$ cm. Oh! That's the mistake. I used sine instead of tangent.

Step1 (corrected): Find pyramid height

$\tan(30^\circ) = \frac{h}{AC}$, so $h = AC \times \tan(30^\circ) = 4\sqrt{3} \times \frac{1}{\sqrt{3}} = 4$ cm

Step2 (corrected): Apply volume formula

Volume $V = \frac{1}{3} \times \text{base area} \times h$
$V = \frac{1}{3} \times 12\sqrt{3} \times 4$

Step3 (corrected): Calculate volume

$V = \frac{1}{3} \times 48\sqrt{3} = 16\sqrt{3}$ cm³

Answer:

$16\sqrt{3}$ cm³ (Option B)