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ISEE Upper Level Math : Tetrahedrons

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Example Questions

Example Question #1 : How To Find The Length Of An Edge Of A Tetrahedron

A triangular pyramid, or tetrahedron, with volume 100 has four vertices with Cartesian coordinates

(0,0,0), (n,0, 0), (0,10,0), (0, 0, 5)

where n > 0 .

Evaluate .

Possible Answers:

n = 12

n = 9

n = 18

n = 24

n = 6

Correct answer:

n = 12

Explanation:

The tetrahedron is as follows (figure not to scale):

Tetrahedron

This is a triangular pyramid with a right triangle with legs 10 and as its base; the area of the base is

$B = \frac{1}{2} \cdot 10 \cdot n = 5n$

The height of the pyramid is 5, so

$V = \frac{1}{3} \cdot 5n \cdot 5 = \frac{25n}{3}$

Set this equal to 100 to get :

$\frac{25n}{3} = 100$

$n = 100 \cdot \frac{3}{25} = 12$

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Example Question #1 : Tetrahedrons

A triangular pyramid, or tetrahedron, with volume 1,000 has four vertices with Cartesian coordinates

(0,0,0), (n,0, 0), (0,n,0), (0, 0, n)

where n > 0 .

Evaluate .

Possible Answers:

n = 10

$n = 10 \sqrt[3]{12}$

$n = 10 \sqrt[3]{2}$

$n = 10 \sqrt[3]{6}$

$n = 10 \sqrt[3]{3}$

Correct answer:

$n = 10 \sqrt[3]{6}$

Explanation:

The tetrahedron is as follows:

Pyramid

This is a triangular pyramid with a right triangle with two legs of measure as its base; the area of the base is

$B = \frac{1}{2} \cdot n \cdot n = \frac{n^{2}}{2}$

Since the height of the pyramid is also , the volume is

$V = \frac{1}{3} \cdot \frac{n^{2}}{2} \cdot n = \frac{n^{3}}{6}$ .

Set this equal to 1,000:

$\frac{n^{3}}{6} = 1,000$

$n^{3} = 6,000$

$n = \sqrt[3]{6,000} = \sqrt[3]{1,000} \cdot \sqrt[3]{6} = 10 \sqrt[3]{6}$

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Example Question #3 : How To Find The Length Of An Edge Of A Tetrahedron

A triangular pyramid, or tetrahedron, with volume 240 has four vertices with Cartesian coordinates

(0,0,0), (n,0, 0), (0,n,0), (0, 0, 24)

where n > 0 .

Evaluate .

Possible Answers:

$n= 3\sqrt{10}$

$n = \sqrt{30}$

$n = 2 \sqrt{30}$

$n = 2 \sqrt{15}$

$n= 6\sqrt{10}$

Correct answer:

$n = 2 \sqrt{15}$

Explanation:

The tetrahedron is as follows (figure not to scale):

Tetrahedron

This is a triangular pyramid with a right triangle with two legs of measure as its base; the area of the base is

$B = \frac{1}{2} \cdot n \cdot n = \frac{n^{2}}{2}$

The height of the pyramid is 24, so the volume is

$V = \frac{1}{3} \cdot \frac{n^{2}}{2} \cdot 24= 4 n^{2}$

Set this equal to 240 to get :

$4 n^{2} = 240$

$n^{2} = 60$

$n = \sqrt{60} = \sqrt{4} \cdot \sqrt{15} = 2 \sqrt{15}$

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Example Question #2 : Tetrahedrons

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates (0,0,0), (18,0,0), (9, 18, 0), (9,9, 9) .

Give its volume.

Possible Answers:

486

1,458

729

243

972

Correct answer:

486

Explanation:

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

Three of the vertices - (0,0,0), (18,0,0), (9, 18, 0) - are on the -plane, and can be seen as the vertices of the triangular base. This triangle, as seen below, is isosceles:

Thingy

Its base and height are both 18, so its area is

$B = \frac{1}{2} \cdot 18 \cdot 18 = 162$

The fourth vertex is off the -plane; its perpendicular distance to the aforementioned face is its -coordinate, 9, so this is the height of the pyramid. The volume of the pyramid is

$V = \frac{1}{3} \cdot 162 \cdot 9 = 486$

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Example Question #3 : Tetrahedrons

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates (0,0,-N), (12,0,-N), (6, 15, -N), ( 6, 10, N) , where N > 0 .

Give its volume in terms of .

Possible Answers:

120N

90N

45N

30N

60N

Correct answer:

60N

Explanation:

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

Three of the vertices - (0,0,-N), (12,0,-N), (6, 15, -N) - are on the horizontal plane xy = -N , and can be seen as the vertices of the triangular base. This triangle, as seen below, is isosceles:

Thingy

Its base is 12 and its height is 15, so its area is

$B = \frac{1}{2} \cdot 12 \cdot 15 = 90$

The fourth vertex is off this plane; its perpendicular (vertical) distance to the aforementioned face is the difference between the -coordinates, N - (-N) = 2N , so this is the height of the pyramid. The volume of the pyramid is

$V = \frac{1}{3} \cdot 90 \cdot 2N = 60N$

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Example Question #4 : Tetrahedrons

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates (0,0,-3), (20,0,-3), (10, 9, -3), ( 10, 6, 3) .

Give its volume in terms of .

Possible Answers:

The correct answer is not among the other choices.

180

270

Correct answer:

180

Explanation:

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

Three of the vertices - (0,0,-3), (20,0,-3), (10, 9, -3) - are on the horizontal plane xy = -3 , and can be seen as the vertices of the triangular base. This triangle, as seen below, is isosceles (drawing not to scale):

Thingy

Its base is 20 and its height is 9, so its area is

$B = \frac{1}{2} \cdot 20 \cdot 9= 90$

The fourth vertex is off this plane; its perpendicular (vertical) distance to the aforementioned face is the difference between the -coordinates, 3 - (-3) = 6 , so this is the height of the pyramid. The volume of the pyramid is

$V = \frac{1}{3} \cdot 90 \cdot 6 = 180$

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Example Question #4 : Tetrahedrons

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates (0,0,0), (12,0,0), (0,12,0), (0,0, 12) .

What is the volume of this tetrahedron?

Possible Answers:

144

864

192

288

576

Correct answer:

288

Explanation:

The tetrahedron looks like this:

Tetrahedron

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

This is a triangular pyramid, so look at $\Delta OCB$ as its base; the area of the base is half the product of its legs, or

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

The volume of the tetrahedron, it being essentially a pyramid, is one third the product of its base and its height, the latter of which is 12. Therefore,

$V = \frac{1}{3} \cdot 12 \cdot 72 = 288$ .

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Example Question #5 : Tetrahedrons

Thingy

Above is the base of a triangular pyramid, which is equilateral. AM = 30 , and the pyramid has height 30. What is the volume of the pyramid?

Possible Answers:

$2,000\sqrt{3}$

3,000

$3,000\sqrt{3}$

$4,500\sqrt{3}$

Correct answer:

$3,000\sqrt{3}$

Explanation:

Altitude $\overline{AM}$ divides $\Delta ABC$ into two 30-60-90 triangles.

By the 30-60-90 Theorem, $AM = CM \sqrt{3}$ , or

$CM = \frac{AM}{\sqrt{3}}$

$CM = \frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{30\sqrt{3}}{3}= 10\sqrt{3}$

is the midpoint of $\overline{BC}$ , so $BC = 2 \cdot CM = 2 \cdot 10\sqrt{3}= 20 \sqrt{3}$

The area of the triangular base is half the product of its base and its height:

$B = \frac{1}{2} \cdot 30 \cdot 20 \sqrt{3}= 300 \sqrt{3}$

The volume of the pyramid is one third the product of this area and the height of the pyramid:

$V = \frac{1}{3} \cdot 30 \cdot 300 \sqrt{3} = 3,000\sqrt{3}$

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Example Question #1 : Tetrahedrons

A regular tetrahedron has edges of length 4. What is its surface area?

Possible Answers:

$16 \sqrt{3}$

$16\sqrt{2}$

$8\sqrt{6}$

Correct answer:

$16 \sqrt{3}$

Explanation:

A regular tetrahedron has four faces, each of which is an equilateral triangle. Therefore, its surface area, given sidelength , is

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

Substitute s = 4 :

$A = s^{2} \sqrt{3} = 4^{2} \cdot \sqrt{3} = 16 \sqrt{3}$

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Example Question #7 : Tetrahedrons

A regular tetrahedron comprises four faces, each of which is an equilateral triangle. Each edge has length 16. What is its surface area?

Possible Answers:

$256 \sqrt{2}$

$682\frac{2}{3}$

$341\frac{1}{3}$

256

$256 \sqrt{3}$

Correct answer:

$256 \sqrt{3}$

Explanation:

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

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

Substitute edge length 16 for :

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

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

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

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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : Tetrahedrons

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #1 : How To Find The Length Of An Edge Of A Tetrahedron

Example Question #1 : Tetrahedrons

Example Question #3 : How To Find The Length Of An Edge Of A Tetrahedron

Example Question #2 : Tetrahedrons

Example Question #3 : Tetrahedrons

Example Question #4 : Tetrahedrons

Example Question #4 : Tetrahedrons

Example Question #5 : Tetrahedrons

Example Question #1 : Tetrahedrons

Example Question #7 : Tetrahedrons

All ISEE Upper Level Math Resources

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