Q27. - Geometry

Question

Use trigonometry to find the volume of the regular pyramid below to the near cubic unit.

Step-by-Step Solution

Answer

$\text{Volume}=66{\text{\hspace{0.17em}\hspace{0.17em}unit}}^3$

Step 1. Given information.

The regular pyramid is given below.

Step 2. Explanation.

The formula for the volume of pyramid is the one third of the product of base area and height.

      $\text{V}=\frac{1}{3}\text{B}h$       

 

Consider the tangent of the angle$50°$ for the height of the hexagonal pyramid.

 $\begin{array}{c}h=\left(\text{base\hspace{0.17em}edge}\right)\times \tan 50°\\ =4\times \tan 50°\\ =4\tan 50°\end{array}$

 

Simplify the base area of hexagonal triangle with side4 .

$\begin{array}{c}\text{B}=\frac{3\sqrt{3}}{2}{a}^2\\ =\frac{3\sqrt{3}}{2}{\left(4\right)}^2\\ =\frac{3\sqrt{3}}{2}\left(16\right)\\ =24\sqrt{3}\end{array}$

Step 3. Determine The volume.

Substitute$24\sqrt{3}$ for B and$4\tan 50°$ for h in the formula for volume of pyramid.

$\begin{array}{c}\text{V}=\frac{1}{3}\text{B}h\\ =\frac{1}{3}\left(24\sqrt{3}\right)(4\tan 50°)\\ =32\sqrt{3}\tan 50°\end{array}$

 

Substitute$1.191$ for$\tan 50°$ in the formula for volume.

 $\begin{array}{c}\text{V}=32\sqrt{3}\tan 50°\\ =32\times \sqrt{3}\times 1.191\\ =66.01\end{array}$

 

Therefore, the volume of the regular pyramid to the nearest cubic unit is equal to 66.