A solid right pyramid with a regular triangular base of side 16 cm is completely submerged into a vertical cylindrical container holding water. The base radius of the cylinder is 8 cm. Prior to placing the pyramid in the container, the depth of the water is 15 cm. After the pyramid is fully underwater, the water level rises to 18.5 cm. Determine the vertical height of the pyramid.

The correct answer is:

$\frac{7\pi \sqrt{3}}{2}$ cm

Step 1: Calculate the volume of the displaced water.
When a solid object is completely submerged in a liquid inside a cylindrical container, the volume of the submerged object equals the volume of water displaced.

Given parameters:
Base radius of the cylindrical container, r = 8 cm
Initial depth of water = 15 cm
Final depth of water = 18.5 cm
Rise in water level, Δh = 18.5 - 15 = 3.5 cm

The volume of displaced water (V) is equal to the volume of a cylinder with height Δh:

$V=\pi r^2\Delta h$

Substituting r = 8 cm and Δh = 3.5 cm:

$V=\pi \times 8^2\times 3.5$

$V=\pi \times 64\times \frac{7}{2}$

$V=224\pi {\text{cm}}^3$

Step 2: Calculate the base area of the pyramid.
The base of the solid right pyramid is a regular triangle (an equilateral triangle) with side length a = 16 cm.

The area (A) of an equilateral triangle is given by the formula:

$A=\frac{\sqrt{3}}{4}{a}^2$

Substituting a = 16 cm into the formula:

$A=\frac{\sqrt{3}}{4}\times {16}^2$

$A=\frac{\sqrt{3}}{4}\times 256$

$A=64\sqrt{3}{\text{cm}}^2$

Step 3: Determine the vertical height of the pyramid.
The volume of a pyramid is related to its base area A and vertical height h by:

$V_{\text{pyramid}}=\frac{1}{3}\times A\times h$

Equating the volume of the pyramid to the volume of the displaced water:

$\frac{1}{3}\times 64\sqrt{3}\times h=224\pi$

Solving for vertical height h:

$h=\frac{224\pi \times 3}{64\sqrt{3}}$

$h=\frac{7\pi \times 3}{2\sqrt{3}}$

Rationalizing the denominator by multiplying both the numerator and denominator by square root of 3:

$h=\frac{7\pi \times 3\sqrt{3}}{2\times 3}$

$h=\frac{7\pi \sqrt{3}}{2}\text{cm}$

Thus, the vertical height of the pyramid is:

$\frac{7\pi \sqrt{3}}{2}$ cm