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 $12\sqrt{3}$ $\mathrm{cm}^2$. what is the volume of the pyramid? \bigcirc $12\sqrt{3}$ $\mathrm{cm}^3$ \bigcirc $16\sqrt{3}$ $\mathrm{cm}^3$ \bigcirc $24\sqrt{3}$ $\mathrm{cm}^3$ \bigcirc $32\sqrt{3}$ $\mathrm{cm}^3$

Explanation:

Step1: Recall the volume formula for a pyramid

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

Step2: Determine the height of the pyramid

We are given a right triangle (triangle \( ABC \)) with angle \( 30^\circ \), hypotenuse? Wait, no, the horizontal side \( AC \) is \( 4\sqrt{3} \) cm, and we can find the height \( h \) (the vertical side from \( B \) to \( AC \)) using trigonometry. In a right triangle, \( \sin(30^\circ)=\frac{h}{AB} \)? Wait, no, actually, looking at the diagram, the angle at \( A \) is \( 30^\circ \), and the adjacent side to the \( 30^\circ \) angle is \( 4\sqrt{3} \) cm? Wait, no, the height \( h \) (the height of the pyramid, which is the perpendicular distance from the apex to the base plane) can be found using the right triangle where the angle is \( 30^\circ \), and the adjacent side (the horizontal segment) is \( 4\sqrt{3} \) cm? Wait, no, actually, in the right triangle, \( \sin(30^\circ)=\frac{h}{ \text{hypotenuse}} \)? Wait, no, let's re - examine. The height \( h \) of the pyramid (the vertical height) can be found using the right triangle with angle \( 30^\circ \) and the side adjacent to the \( 30^\circ \) angle? Wait, no, the formula for the height \( h \) in the right triangle: if we have a right triangle with angle \( 30^\circ \), and the side opposite to \( 30^\circ \) is the height \( h \), and the hypotenuse? Wait, no, the length of the side adjacent to the \( 30^\circ \) angle is \( 4\sqrt{3} \) cm? Wait, no, actually, the height \( h \) (the height of the pyramid) can be calculated using \( \sin(30^\circ)=\frac{h}{ \text{the side}} \)? Wait, no, let's use the definition of sine: \( \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}} \). Wait, in the diagram, the angle at \( A \) is \( 30^\circ \), and the side \( AC = 4\sqrt{3} \) cm. Wait, no, the height \( h \) (the height of the pyramid) is the length of \( BC \) (since \( BC \) is perpendicular to \( AC \), so \( BC \) is the height of the pyramid). In triangle \( ABC \), \( \angle A = 30^\circ \), and we can consider that the height \( h = BC \), and we can use \( \tan(30^\circ)=\frac{h}{AC} \)? Wait, no, \( \sin(30^\circ)=\frac{h}{AB} \), but we don't know \( AB \). Wait, no, actually, in a right triangle, if the angle is \( 30^\circ \), and the side adjacent to the \( 30^\circ \) angle is \( 4\sqrt{3} \) cm, then the opposite side (height \( h \)) is given by \( h = 4\sqrt{3}\times\sin(30^\circ) \)? Wait, no, \( \sin(30^\circ)=\frac{1}{2} \), and if we consider that the height \( h \) is opposite the \( 30^\circ \) angle, and the adjacent side is \( 4\sqrt{3} \) cm, then \( \tan(30^\circ)=\frac{h}{4\sqrt{3}} \)? Wait, no, I think I made a mistake. Let's start over.

The volume formula is \( V=\frac{1}{3}Bh \), where \( B = 12\sqrt{3}\space cm^{2} \) (given area of the base). We need to find \( h \).

Looking at the right triangle (the one with the \( 30^\circ \) angle), the height \( h \) of the pyramid (the perpendicular height) can be found using the fact that in a right triangle, \( \sin(30^\circ)=\frac{h}{ \text{the side}} \). Wait, the side adjacent to the \( 30^\circ \) angle is \( 4\sqrt{3} \) cm? No, actually, the height \( h \) is calculated as follows: in the right triangle, \( \sin(30^{\circ})=\frac{h}{ \text{the length of the side from }A\text{ to the point above }C} \)? Wait, no, the correct way is: we know that in a right triangle, if the angle is \( 30^\circ \), and the hypotenuse is not given, but we can use the fact that \( \sin(30^\circ)=\frac{1}{2} \), and the side opposite to \( 30^\circ \) is \( h \), and the adjacent side is \( 4\sqrt{3} \) cm? No, let's use trigonometry correctly. Let's assume that the height \( h \) (the height of the pyramid) is the length of \( BC \). In triangle \( ABC \), \( \angle A = 30^\circ \), and \( AC = 4\sqrt{3} \) cm. We can use the formula \( \tan(30^\circ)=\frac{h}{AC} \)? No, \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \). So \( \tan(30^\circ)=\frac{h}{4\sqrt{3}} \), but \( \tan(30^\circ)=\frac{1}{\sqrt{3}} \), so \( h = 4\sqrt{3}\times\frac{1}{\sqrt{3}} = 4 \)? Wait, no, that's not right. Wait, \( \sin(30^\circ)=\frac{h}{ \text{the hypotenuse}} \), but we don't know the hypotenuse. Wait, maybe I got the angle wrong. Wait, the height \( h \) of the pyramid can be found using \( \sin(30^\circ)=\frac{h}{ \text{the side}} \), where the side is \( 4\sqrt{3} \) cm? No, \( \sin(30^\circ)=\frac{1}{2} \), so if we have a right triangle with angle \( 30^\circ \), and the side opposite to \( 30^\circ \) is \( h \), and the hypotenuse is \( 4\sqrt{3} \) cm, then \( h = 4\sqrt{3}\times\sin(30^\circ)=4\sqrt{3}\times\frac{1}{2}=2\sqrt{3} \)? No, that can't be. Wait, no, the correct approach: the volume of a pyramid is \( V = \frac{1}{3}Bh \), where \( B \) is the base area (\( 12\sqrt{3}\space cm^{2} \)) and \( h \) is the height. To find \( h \), we look at the right triangle with angle \( 30^\circ \). The height \( h \) (the perpendicular height of the pyramid) is related to the \( 4\sqrt{3} \) cm side. Wait, in a right triangle, if the angle is \( 30^\circ \), the side opposite \( 30^\circ \) is half the hypotenuse, but here we have a right triangle where the adjacent side to \( 30^\circ \) is \( 4\sqrt{3} \) cm? No, I think I made a mistake. Let's re - express:

We know that the base area \( B = 12\sqrt{3}\space cm^{2} \). We need to find the height \( h \) of the pyramid. From the diagram, we have a right triangle with angle \( 30^\circ \), and the side adjacent to the \( 30^\circ \) angle is \( 4\sqrt{3} \) cm? No, the height \( h \) (the height of the pyramid) can be found using \( \sin(30^\circ)=\frac{h}{ \text{the length of the side}} \). Wait, the length of the side (the hypotenuse of the right triangle) is not given, but we can use the fact that in a right triangle, \( \sin(30^\circ)=\frac{1}{2} \), so if we consider that the height \( h \) is opposite the \( 30^\circ \) angle, and the adjacent side is \( 4\sqrt{3} \) cm, then \( \tan(30^\circ)=\frac{h}{4\sqrt{3}} \), but \( \tan(30^\circ)=\frac{1}{\sqrt{3}} \), so \( h = 4\sqrt{3}\times\frac{1}{\sqrt{3}} = 4 \)? Wait, no, that would be if we use tangent. Wait, no, let's use sine. If the angle is \( 30^\circ \), and the hypotenuse is \( 4\sqrt{3} \) cm, then \( h=\sin(30^\circ)\times4\sqrt{3}=\frac{1}{2}\times4\sqrt{3}=2\sqrt{3} \)? No, this is confusing. Wait, maybe the height \( h \) is calculated as follows: in the right triangle, the angle is \( 30^\circ \), and the side opposite to \( 30^\circ \) is \( h \), and the side adjacent is \( 4\sqrt{3} \) cm. Wait, no, the correct way is: the height \( h \) of the pyramid (the perpendicular height) is equal to \( 4\sqrt{3}\times\sin(30^\circ) \)? Wait, \( \sin(30^\circ)=\frac{1}{2} \), so \( h = 4\sqrt{3}\times\frac{1}{2}=2\sqrt{3} \)? No, that can't be. Wait, maybe I misread the diagram. Wait, the length of \( AC \) is \( 4\sqrt{3} \) cm, and the angle at \( A \) is \( 30^\circ \), and \( BC \) is perpendicular to \( AC \), so triangle \( ABC \) is a right triangle with \( \angle C = 90^\circ \), \( \angle A = 30^\circ \), and \( AC = 4\sqrt{3} \) cm. Then, \( \tan(30^\circ)=\frac{BC}{AC} \), so \( BC = AC\times\tan(30^\circ) \). Since \( \tan(30^\circ)=\frac{1}{\sqrt{3}} \), then \( BC = 4\sqrt{3}\times\frac{1}{\sqrt{3}} = 4 \) cm. Ah, that's it! So the height \( h \) of the pyramid is \( 4 \) cm.

Step3: Calculate the volume of the pyramid

Now that we have \( B = 12\sqrt{3}\space cm^{2} \) and \( h = 4\space cm \), we use the volume formula \( V=\frac{1}{3}Bh \).

Substitute the values: \( V=\frac{1}{3}\times12\sqrt{3}\times4 \)

First, calculate \( \frac{1}{3}\times12\sqrt{3}=4\sqrt{3} \)

Then, multiply by \( 4 \): \( 4\sqrt{3}\times4 = 16\sqrt{3}\space cm^{3} \)

Answer:

\( 16\sqrt{3}\space cm^{3} \) (corresponding to the option \( 16\sqrt{3}\space cm^{3} \))