Tetrahedrons - ISEE Upper Level Mathematics Achievement

Card 0 of 12

Question

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

where .

Evaluate .

Answer

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

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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Question

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

where .

Evaluate .

Answer

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

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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Question

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

where .

Evaluate .

Answer

The tetrahedron is as follows:

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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Question

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

Answer

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 :

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

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Question

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

What is the surface area of this tetrahedron?

Answer

The tetrahedron looks like this:

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

Three of the surfaces are right triangles with two legs of length 12, so the area of each is

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

The fourth surface, $\Delta ABC$ , has three edges each of which is the hypotenuse of an isosceles right triangle with legs 12, so each has length $12\sqrt{2}$ by the 45-45-90 Theorem. That makes this triangle equilateral, so its area is'

$\frac{(12 \sqrt{2}) ^{2} \sqrt{3}}{4}= \frac{144 \cdot 2 \cdot \sqrt{3}}{4} = 72 \sqrt{3}$

The surface area is therefore

$3 \times 72 + 72 \sqrt{3} = 216+ 72 \sqrt{3}$ .

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Question

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

Answer

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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Question

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

In terms of , give the surface area of this tetrahedron.

Answer

The tetrahedron looks like this:

is the origin and are the other three points, which are units away from the origin, each along one of the three (perpendicular) axes.

Three of the surfaces are right triangles with two legs of length 12, so the area of each is

$A = \frac{1}{2} \cdot n\cdot n = \frac{n^{2}}{2}$ .

The fourth surface, $\Delta ABC$ , has three edges each of which is the hypotenuse of an isosceles right triangle with legs , so each has length $n\sqrt{2}$ by the 45-45-90 Theorem. That makes this triangle equilateral, so its area is'

$\frac{(n \sqrt{2}) ^{2} \sqrt{3}}{4}= \frac{n^{2}\cdot 2 \cdot \sqrt{3}}{4} = \frac{n^{2} \sqrt{3}}{2}$

The surface area is therefore

$3 \cdot \frac{n^{2}}{2} + \frac{n^{2} \sqrt{3}}{2} = \frac{3n^{2}+ n^{2} \sqrt{3}}{2}$ .

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Question

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

What is the volume of this tetrahedron?

Answer

The tetrahedron looks like this:

is the origin and 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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Question

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

Answer

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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Question

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

Give its volume in terms of .

Answer

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

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

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, , 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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Question

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

Give its volume.

Answer

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:

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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Question

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

Give its volume in terms of .

Answer

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

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

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, , 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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