The area of each face of a regular tetrahedron, that face being an equilateral triangle, is 

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

Substitute edge length 16 for $\displaystyle s$:

$\displaystyle A = \frac{16^{2}\sqrt{3}}{4} = \frac{256\sqrt{3}}{4} = 64 \sqrt{3}$

The tetrahedron has four faces, so the total surface area is 

$\displaystyle SA = 4 \cdot 64 \sqrt{3} = 256 \sqrt{3}$

The tetrahedron looks like this:

$\displaystyle O$ is the origin and $\displaystyle A,B,C$ are the other three points, which are each twelve units away from the origin on one of the three (mutually perpendicular) axes.

Three of the surfaces are right triangles with two legs of length 12, so the area of each is 

$\displaystyle A = \frac{1}{2} \cdot 12 \cdot 12 = 72$.

The fourth surface, $\displaystyle \Delta ABC$, has three edges each of which is the hypotenuse of an isosceles right triangle with legs 12, so each has length $\displaystyle 12\sqrt{2}$ by the 45-45-90 Theorem. That makes this triangle equilateral, so its area is'

$\displaystyle \frac{(12 \sqrt{2}) ^{2} \sqrt{3}}{4}= \frac{144 \cdot 2 \cdot \sqrt{3}}{4} = 72 \sqrt{3}$

The surface area is therefore

$\displaystyle 3 \times 72 + 72 \sqrt{3} = 216+ 72 \sqrt{3}$.

The tetrahedron looks like this:

$\displaystyle O$ is the origin and $\displaystyle A,B,C$ are the other three points, which are $\displaystyle n$ units away from the origin, each along one of the three (perpendicular) axes.

Three of the surfaces are right triangles with two legs of length 12, so the area of each is 

$\displaystyle A = \frac{1}{2} \cdot n\cdot n = \frac{n^{2}}{2}$.

The fourth surface, $\displaystyle \Delta ABC$, has three edges each of which is the hypotenuse of an isosceles right triangle with legs $\displaystyle n$, so each has length $\displaystyle n\sqrt{2}$ by the 45-45-90 Theorem. That makes this triangle equilateral, so its area is'

$\displaystyle \frac{(n \sqrt{2}) ^{2} \sqrt{3}}{4}= \frac{n^{2}\cdot 2 \cdot \sqrt{3}}{4} = \frac{n^{2} \sqrt{3}}{2}$

The surface area is therefore

$\displaystyle 3 \cdot \frac{n^{2}}{2} + \frac{n^{2} \sqrt{3}}{2} = \frac{3n^{2}+ n^{2} \sqrt{3}}{2}$.

There is a perfectly spherical weather balloon with a surface area of  $\displaystyle 900 \pi ft^2$, what is its diameter?

Begin with the formula for surface area of a sphere:

$\displaystyle SA_{sphere}=4 \pi r^2$

Now, set it equal to the given surface area and solve for r:

$\displaystyle 900 \pi ft^2=4 \pi r^2$

First divide both sides by $\displaystyle 4\pi$.

$\displaystyle 225ft^2=r^2$

Then square root both sides to get our radius:

$\displaystyle r=15ft$

Now, because the question is asking for our diameter and not our radius, we need to double our radius to get our answer:

$\displaystyle d=2r=2*15ft=30 ft$

A wooden ball has a surface area of $\displaystyle 900\pi m^2$.

What is its radius?

Begin with the formula for surface area of a sphere:

$\displaystyle SA_{sphere}=4\pi r^2$

Now, plug in our surface area and solve with algebra:

$\displaystyle 900 \pi m^2=4\pi r^2$

Get rid of the pi

$\displaystyle 900 m^2=4 r^2$

Divide by 4

$\displaystyle 225 m^2= r^2$

Square root both sides to get our answer:

$\displaystyle r=15m$

There is a perfectly spherical weather balloon with a surface area of  $\displaystyle 900 \pi ft^2$, what is its radius?

Begin with the formula for surface area of a sphere:

$\displaystyle SA_{sphere}=4 \pi r^2$

Now, set it equal to the given surface area and solve for r:

$\displaystyle 900 \pi ft^2=4 \pi r^2$

First divide both sides by $\displaystyle 4\pi$.

$\displaystyle 225ft^2=r^2$

Then square root both sides to get our answer:

$\displaystyle r=15ft$

20 feet = $\displaystyle 20 \times 12 = 240$ inches, the diameter of the tank; half of this, or 120 inches, is the radius. Set $\displaystyle r = 120$, substitute in the volume formula, and solve for $\displaystyle V$:

$\displaystyle V = \frac{4}{3} \pi r^{3}$

$\displaystyle V = \frac{4}{3} \pi \cdot 120^{3}$

$\displaystyle V = \frac{4}{3} \pi \cdot 1,728,000$

$\displaystyle V = 2,304,000 \pi$ cubic inches

The diameter of 3 meters is equal to $\displaystyle 3 \times 100 = 300$ centimeters; the radius is half this, or 150 centimeters. Substitute $\displaystyle r = 150$ in the volume formula:

$\displaystyle V= \frac{4}{3} \pi r^{3}$

$\displaystyle V= \frac{4}{3} \cdot \pi \cdot 150^{3}$

$\displaystyle V= \frac{4}{3} \cdot 150 \cdot 150 \cdot 150\cdot \pi$

$\displaystyle V= \frac{4}{3} \cdot 150 \cdot 150 \cdot 150\cdot \pi$

$\displaystyle V= 4,500,000\pi$ cubic centimeters

A spherical buoy has a radius of 5 meters. What is the volume of the buoy?

To find the volume of a sphere, use the following formula:

$\displaystyle V_{sphere}=\frac{4}{3}\pi r^3$

All we have to do is plug in 5 meters and simplify:

$\displaystyle V_{sphere}=\frac{4}{3}\pi (5m)^3=\frac{4}{3}\pi*125m^3=166\frac{2}{3}\pi m^3$

You have a ball with a radius of  12 cm, what is its volume?

The volume of a sphere can be found via the following formula:

$\displaystyle V_{sphere}=\frac{4}{3}\pi r^3$

We know our radius, so all we need to do is plug in and simplify:

$\displaystyle V_{sphere}=\frac{4}{3}\pi (12cm)^3=2304 \pi cm^3$

So we have our answer:

$\displaystyle 2304 \pi cm^3$