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

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Example Questions

Example Question #1 : Tetrahedrons

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

\displaystyle (0,0,0), (n,0, 0), (0,10,0), (0, 0, 5)

where \displaystyle n > 0.

Evaluate \displaystyle n

Possible Answers:

\displaystyle n = 6

\displaystyle n = 24

\displaystyle n = 18

\displaystyle n = 12

\displaystyle n = 9

Correct answer:

\displaystyle 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 \displaystyle n as its base; the area of the base is 

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

The height of the pyramid is 5, so

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

Set this equal to 100 to get \displaystyle n:

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

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

 

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

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

\displaystyle (0,0,0), (n,0, 0), (0,n,0), (0, 0, n)

where \displaystyle n > 0.

Evaluate \displaystyle n

Possible Answers:

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

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

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

\displaystyle n = 10

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

Correct answer:

\displaystyle 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 \displaystyle n as its base; the area of the base is 

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

Since the height of the pyramid is also \displaystyle n, the volume is

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

Set this equal to 1,000:

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

\displaystyle n^{3} = 6,000

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

Example Question #2 : Tetrahedrons

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

\displaystyle (0,0,0), (n,0, 0), (0,n,0), (0, 0, 24)

where \displaystyle n > 0.

Evaluate \displaystyle n

Possible Answers:

\displaystyle n= 6\sqrt{10}

\displaystyle n = 2 \sqrt{15}

\displaystyle n = \sqrt{30}

\displaystyle n = 2 \sqrt{30}

\displaystyle n= 3\sqrt{10}

Correct answer:

\displaystyle 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 \displaystyle n as its base; the area of the base is 

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

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

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

Set this equal to 240 to get \displaystyle n:

\displaystyle 4 n^{2} = 240

\displaystyle n^{2} = 60

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

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