What is the volume of a regular tetrahedron with edges of ?

$\small \frac{36}{\sqrt{2}} cm^3$

$\small \frac{216}{\sqrt{2}} cm^3$

$\small \frac{36}{\sqrt{2}} cm^3$

$\small 6\sqrt{2} cm^3$

$\small \sqrt{2} cm^3$

Explanation

The volume of a tetrahedron is found with the formula:

$\small V= \frac{a^3}{6\sqrt{2}}$ ,

where $\small a$ is the length of the edges.

When $\small a=6$ ,

$\small V=\frac{36}{\sqrt{2}}$ .

A circular swimming pool has diameter 40 meters and depth meters throughout. Which of the following expressions gives the amount of water it holds, in cubic meters?

$400 \pi t$

$1,600 \pi t$

$40 \pi t$

$80 \pi t$

$160 \pi t$

Explanation

The pool can be seen as a cylinder with depth (or height) , and a base with diameter 40 m - and radius half this, or . The capacity of the pool is the volume of this cylinder, which is

$V = \pi r^{2}h = \pi \cdot 20^{2}\cdot t = 400 \pi t$

The radius and the height of a cylinder are equal. If the volume of the cylinder is , what is the diameter of the cylinder?

Explanation

Recall how to find the volume of a cylinder:

$\text{Volume}=\pi r^2 h$

Since we know that the radius and the height are equal, we can rewrite the equation:

$\text{Volume}=\pi r^2 (r)=\pi r^3$

Using the given volume, find the length of the radius.

$13=\pi r^3$

Since the question asks you to find the diameter, multiply the radius by two.

The bottom surface of a rectangular prism has area 100; the right surface has area 200; the rear surface has area 300. Give the volume of the prism (nearest whole unit), if applicable.

Explanation

Let the dimensions of the prism be , , and .

Then, , , and .

From the first and last equations, dividing both sides, we get

$\frac{WH}{LW} = \frac{300}{100}$

$\frac{H}{L} = 3$

Along with the second equation, multiply both sides:

$\frac{H}{L} \cdot LH = 3 \cdot 200$

$H ^{2}= 600$

Taking the square root of both sides and simplifying, we get

$H = \sqrt{600} = \sqrt{100} \cdot \sqrt{6} = 10 \sqrt{6}$

Now, substituting and solving for the other two dimensions:

$L \cdot 10 \sqrt{6} = 200$

$L = \frac{200}{10 \sqrt{6}} = \frac{20}{\sqrt{6}}$

$W \cdot 10 \sqrt{6} = 300$

$W = \frac{300}{10 \sqrt{6}} = \frac{30 }{\sqrt{6}}$

Now, multiply the three dimensions to obtain the volume:

$= \frac{20}{\sqrt{6}} \cdot 10 \sqrt{6} \cdot \frac{30 }{\sqrt{6}}$

$= \frac{20}{\sqrt{6}} \cdot\frac{ 10 \sqrt{6}}{1} \cdot \frac{30 }{\sqrt{6}}$

$= \frac{ 6,000 \sqrt{6}} {6}$

$= 1,000 \sqrt{6}$

$\approx 1,000 \cdot 2.449$

$\approx 2,449$

The width of a box is two-thirds its height and three-fifths its length. The volume of the box is 6 cubic meters. To the nearest centimeter, give the width of the box.

$134 \textrm{ cm}$

$201 \textrm{ cm}$

$223 \textrm{ cm}$

$246 \textrm{ cm}$

$164 \textrm{ cm}$

Explanation

Call , , and the length, width, and height of the crate.

The width is two-thirds the height, so

$W = \frac{2}{3} H$ .

Equivalently,

$\frac{3} {2} \cdot \frac{2}{3} H = \frac{3} {2} W$

$H = \frac{3} {2} W$

The width is three-fifths the length, so

$W = \frac{3}{5} L$ .

Equivalently,

$\frac{5} {3} \cdot \frac{3}{5} L = \frac{5} {3} \cdot W$

$L = \frac{5} {3 } W$

The dimensions of the crate in terms of are , $\frac{3} {2} W$ , and $\frac{5} {3 } W$ . The volume is their product:

Substitute:

$\frac{5} {3 } W \cdot W \cdot \frac{3} {2} W = V$

$\frac{5} {3 } \cdot \frac{3} {2} \cdot W \cdot W \cdot W = 6$

$\frac{5} {2} \cdot W ^{3}= 6$

$\frac{2} {5} \cdot \frac{5} {2} \cdot W ^{3}= \frac{2} {5} \cdot 6$

$W ^{3}= \frac{12} {5}$

Taking the cube root of both sides:

$W =\sqrt[3]{ \frac{12} {5}}$

$W \approx 1.339$ meters.

Since one meter comprises 100 centimeters, multiply by 100 to convert to centimeters:

$1.339 \cdot 100 = 133.9$ centimeters,

which rounds to 134 centimeters.

Cone example

Figure not drawn to scale

If the volume of the cone above is 47.12 ft3, what is the radius of the base?

3 ft

2 ft

3.5ft

4 ft

5 ft

Explanation

Because we have been given the volume of the cone and have been asked to find the radius of the base of the cone, we must work backwards using the volume formula.

Cone example

$\small Volume= {\frac{1}{3}}\pi r{^{2}}h$

$\small \small 47.12={\frac{1}{3}}\pi r{^{2}}(5)={\frac{5}{3}}\pi r{^{2}}$

$\small \left({\frac{3}{5}} \right)47.12=\left({\frac{3}{5}} \right )\left({\frac{5}{3}} \right )\pi r{^{2}}$

$\small \frac{28.27}{\pi}=r{^{2}}$

$\small 9.00=r{^{2}}$

$\small r=3$

The radius of the base of the cone is 3 ft.

What is the volume of the following tetrahedron? Assume the figure is a regular tetrahedron.

$\frac{2\sqrt{2}}{3}m^3$

$\4\sqrt{3}m^3$

$\frac{\sqrt{2}}{3}m^3$

$\frac{8\sqrt{2}}{3}m^3$

$\frac{3\sqrt{2}}{2}m^3$

Explanation

A regular tetrahedron is composed of four equilateral triangles. The formula for the volume of a regular tetrahedron is:

$V=\frac{a^3\sqrt{2}}{12}$ , where represents the length of the side.

Plugging in our values we get:

$V=\frac{(2: m)^3\sqrt{2}}{12}$

$V=\frac{8: m^3\sqrt{2}}{12} = \frac{2\sqrt{2}}{3}m^3$

One cubic meter is equal to one thousand liters.

A circular swimming pool is meters in diameter and meters deep throughout. How many liters of water does it hold?

$250\pi d^{2}f$

$500 \pi d^{2}f$

$1,000\pi d^{2}f$

$1,000\pi df$

$500\pi df$

Explanation

The pool can be seen as a cylinder with depth (or height) , and a base with diameter - and, subsequently, radius half this, or $\frac{d}{2}$ . The volume of the pool in cubic meters is

$V = \pi r^{2}h = \pi \left ( \frac{d}{2} \right )^{2} \cdot f = \pi \cdot \frac{d^{2}}{4} \cdot f = \frac{ \pi d^{2}f}{4}$

Multiply this number of cubic meters by 1,000 liters per cubic meter:

$V = \frac{ \pi d^{2}f}{4} \cdot \1,000 = 250\pi d^{2}f$

Pool

The above depicts a rectangular swimming pool for an apartment.

On the left and right edges, the pool is three feet deep; the dashed line at the very center represents the line along which it is eight feet deep. Going from the left to the center, its depth increases uniformly; going from the center to the right, its depth decreases uniformly.

In cubic feet, how much water does the pool hold?

$9,625 \textrm{ ft}^{3}$