How to find the volume of a tetrahedron

Help Questions

SSAT Upper Level Quantitative › How to find the volume of a tetrahedron

Questions 1 - 10

1

Find the volume of a regular tetrahedron that has a side length of .

$\frac{16\sqrt2}{3}$

$32\sqrt2$

$\frac{64}{3\sqrt2}$

$\frac{8\sqrt2}{3}$

Explanation

Use the following formula to find the volume of a regular tetrahedron:

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}4^3$

$\text{Volume}=\frac{64\sqrt2}{12}$

$\text{Volume}=\frac{16\sqrt2}{3}$

2

Find the volume of a regular tetrahedron that has a side length of .

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

$\frac{25\sqrt2}{12}$

$20\sqrt2$

$\frac{5\sqrt2}{6}$

Explanation

Use the following formula to find the volume of a regular tetrahedron:

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}5^3$

$\text{Volume}=\frac{125\sqrt2}{12}$

3

Find the volume of a regular tetrahedron with a side length of .

$18\sqrt2$

$6\sqrt2$

$36\sqrt2$

$24\sqrt2$

Explanation

Use the following formula to find the volume of a regular tetrahedron:

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}6^3$

$\text{Volume}=\frac{216\sqrt2}{12}$

$\text{Volume}=18\sqrt2$

4

Find the volume of a regular tetrahedron with side lengths of .

$\frac{343\sqrt2}{12}$

$\frac{49\sqrt2}{12}$

$\frac{7\sqrt2}{6}$

$18\sqrt2$

Explanation

Use the following formula to find the volume of a regular tetrahedron:

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}7^3$

$\text{Volume}=\frac{343\sqrt2}{12}$

5

Find the volume of a regular tetrahedron with side lengths of .

$\frac{128\sqrt2}{3}$

$\frac{16\sqrt2}{3}$

$2\sqrt2$

$10\sqrt2$

Explanation

Use the following formula to find the volume of a regular tetrahedron:

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}8^3$

$\text{Volume}=\frac{512\sqrt2}{12}$

$\text{Volume}=\frac{128\sqrt2}{3}$

6

Find the volume of a tetrahedron with side lengths of .

$144\sqrt2$

$1728\sqrt2$

$168\sqrt2$

$966\sqrt2$

Explanation

Use the following formula to find the volume of a regular tetrahedron:

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}12^3$

$\text{Volume}=\frac{1728\sqrt2}{12}$

$\text{Volume}=144\sqrt2$

7

Find the volume of a regular tetrahedron with side lengths of .

$\frac{250\sqrt2}{3}$

$\frac{1000\sqrt2}{3}$

$\frac{500\sqrt2}{3}$

$125\sqrt2$

Explanation

Use the following formula to find the volume of a regular tetrahedron:

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}10^3$

$\text{Volume}=\frac{1000\sqrt2}{12}$

$\text{Volume}=\frac{250\sqrt2}{3}$

8

Find the volume of a regular tetrahedron with side lengths of .

$\frac{2x^3\sqrt2}{3}$

$\frac{3x^3\sqrt2}{4}$

$8x^3\sqrt2$

$\frac{4x^3\sqrt2}{3}$

Explanation

Use the following formula to find the volume of a regular tetrahedron:

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}(2x)^3$

$\text{Volume}=\frac{8x^3\sqrt2}{12}$

$\text{Volume}=\frac{2x^3\sqrt2}{3}$

9

Find the volume of a regular tetrahedron with side lengths of .

$\frac{16a^3\sqrt2}{3}$

$\frac{32a^3\sqrt2}{3}$

$16a^3\sqrt2$

$64a^3\sqrt2$

Explanation

Use the following formula to find the volume of a regular tetrahedron:

$\text{Volume}=\frac{\sqrt2}{12}side^3$

Now, plug in the given side length.

$\text{Volume}=\frac{\sqrt2}{12}(4a)^3$

$\text{Volume}=\frac{64a^3\sqrt2}{12}$

$\text{Volume}=\frac{16a^3\sqrt2}{3}$

10

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates .

Give its volume.

Explanation

A tetrahedron is a triangular pyramid and can be looked at as such.

Three of the vertices - - are on the -plane, and can be seen as the vertices of the triangular base. This triangle, as seen below, is isosceles:

Tetrahedron

Its base is 10 and its height is 18, so its area is

$B = \frac{1}{2} \cdot 10 \cdot 18 = 90$

The fourth vertex is off the -plane; its perpendicular distance to the aforementioned face is its -coordinate, 8, so this is the height of the pyramid. The volume of the pyramid is

$V = \frac{1}{3} \cdot 90 \cdot 8 = 240$

Page 1 of 2

Return to subject