Volume

SAT Subject Test in Math I · Learn by Concept

Help Questions

SAT Subject Test in Math I › Volume

1 - 10

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

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

CORRECT

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

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

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

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

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 $\small a=6$ ,

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

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

CORRECT

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.

Find the volume of a tetrahedron with an edge of $\sqrt5$ .

$\frac{25\sqrt10}{12}$

CORRECT

$\frac{125}{6}$

$\frac{5\sqrt3}{9}$

$5\sqrt3$

$\frac{125\sqrt2}{12}$

Explanation

Write the formula the volume of a tetrahedron and substitute in the provided edge length.

$V=\frac{s^3}{6\sqrt 2}= \frac{(\sqrt 5)^3}{6\sqrt 2} = \frac{25\sqrt 5}{6\sqrt2}$

Rationalize the denominator to arrive at the correct answer.

$\frac{25\sqrt5}{6\sqrt2} \cdot \frac{\sqrt2}{\sqrt2} = \frac{25\sqrt 10}{12}$

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

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

CORRECT

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

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

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

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

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:

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

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

What is the volume of a regular tetrahedron with edges of $\small \small \frac{4}{5}cm$ ?

None of the above.

CORRECT

$\small \frac{32}{\sqrt{2}}$ $\small cm^3$

$\small \frac{125}{\sqrt{2}}$ $\small cm^3$

$\small 375\sqrt{2}$

$\small \frac{375}{32\sqrt{2}}$

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 $\small \small a= \frac{4}{5}$

$\small \small V= \left ( \frac{4}{5} ^3\right )\div 6\sqrt{2}$

$\small \small V=\frac{64}{125}\div 6\sqrt{2}$

$\small \small \small V= \frac{64}{125}\times\frac{1}{6\sqrt{2}}=\frac{64}{750\sqrt{2}}=\frac{32}{375\sqrt{2}}$

This answer is not found, so it is "none of the above."

What is the volume of a regular tetrahedron with an edge length of 6?

CORRECT

Explanation

The volume of a tetrahedron can be solved for by using the equation:

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

where is the measurement of the edge of the tetrahedron.

This problem can be quickly solved by substituting 6 in for .

$V=\frac{6^3}{6\sqrt{2}}=\frac{216}{6\sqrt{2}}=\frac{36}{\sqrt{2}}$

${\color{Blue} V=25.5}$

Find the volume of the regular tetrahedron with side length .

CORRECT

Explanation

The formula for the volume of a regular tetrahedron is:

$V = \frac{s^3}{6\sqrt{2}}$

Where is the length of side. Using this formula and the given values, we get:

$V = \frac{5:cm^3}{6\sqrt{2}} = 14.73:cm^3$

What is the volume of a regular tetrahedron with edges of $\small 22$ $\small cm$ ?

$\small \frac{5324}{3\sqrt{2}}$ $\small cm^3$

CORRECT

$\small \frac{5324}{3\sqrt{2}}$ $\small cm^2$

$\small \frac{2662}{3\sqrt{2}}$ $\small cm^3$

$\small \frac{2662}{3\sqrt{2}}$ $\small cm^2$

None of the above.

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 $\small \small a= 22$ ,

$\small \small \small V=\frac{22^3}{6\sqrt{2}}$

$\small V=\frac{10648}{6\sqrt{2}}=\frac{5324}{3\sqrt{2}}$

And, of course, volume should be in cubic measurements!

A circular swimming pool has diameter meters and depth 2 meters throughout. Which of the following expressions give the amount of water it holds, in cubic meters?

$\frac{ \pi k^{2}} {2}$

CORRECT

$2\pi k^{2}$

$\frac{ \pi k^{2}} {4}$

$2\pi k$

$4\pi k$

Explanation

The pool can be seen as a cylinder with diameter - and, subsequently, radius half this, or $\frac{k}{2}$ - and depth, or height, 2. The volume of a cylinder is defined by the formula

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

How is the volume of a regular tetrahedron effected when the length of each edge is doubled?

It is 8 times greater.

CORRECT

It is 4 times greater.

It is doubled.

It increases by 50%.

It cannot be determined by the information given.

Explanation

The volume of a regular tetrahedron is found with the formula $\small \small V_{1}= \frac{a^3}{6\sqrt{2}}$ where $\small a$ is the length of the edges.

The volume of the same tetrahedron when the length of the edges are doubled would be $\small \small \small V_{2}= \frac{(2a)^3}{6\sqrt{2}}$ .

Therefore,

$\small V_{2}=\frac{8a^3}{6\sqrt{2}}=8\times V_{1}$

Return to subject

AI Study CoachYour personal study buddy

Instant TutoringConnect with a tutor now

Request a TutorGet matched with an expert

Live ClassesJoin interactive classes

Question of the DayDaily practice to build your skills

FlashcardsStudy and memorize key concepts

WorksheetsBuild custom practice quizzes

AI SolverStep-by-step problem solutions

Learn by ConceptMaster concepts step by step

Diagnostic TestsAssess your knowledge level

Practice TestsTest your skills with exam questions

GamesLearn through free games