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ISEE Upper Level Math : How to find the length of an edge of a tetrahedron

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Example Questions

Example Question #1 : Tetrahedrons

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 = 6

n = 9

n = 18

n = 12

n = 24

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:

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

n = 10

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

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

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

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 : Tetrahedrons

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:

$n = 2 \sqrt{15}$

$n = \sqrt{30}$

$n= 3\sqrt{10}$

$n= 6\sqrt{10}$

$n = 2 \sqrt{30}$

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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ISEE Upper Level Math : How to find the length of an edge of a tetrahedron

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

All ISEE Upper Level Math Resources

Example Questions

Example Question #1 : Tetrahedrons

Example Question #1 : Tetrahedrons

Example Question #3 : Tetrahedrons

All ISEE Upper Level Math Resources

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