How to find the surface area of a tetrahedron - ISEE Upper Level Quantitative Reasoning

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Question

Which is the greater quantity?

(a) The surface area of a regular tetrahedron with edges of length 1

(b) 2

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 :

$A = s^{2} \sqrt{3} =1^{2} \cdot \sqrt{3} = \sqrt{3}$

$\sqrt{3} \approx 1.7 < 2$ , so (b) is greater.

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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

Which is the greater quantity?

(a) The surface area of a regular tetrahedron with edges of length 1

(b) 2

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 :

$A = s^{2} \sqrt{3} =1^{2} \cdot \sqrt{3} = \sqrt{3}$

$\sqrt{3} \approx 1.7 < 2$ , so (b) is greater.

Compare your answer with the correct one above

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 :

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

Question

Which is the greater quantity?

(a) The surface area of a regular tetrahedron with edges of length 1

(b) 2

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 :

$A = s^{2} \sqrt{3} =1^{2} \cdot \sqrt{3} = \sqrt{3}$

$\sqrt{3} \approx 1.7 < 2$ , so (b) is greater.

Compare your answer with the correct one above

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 :

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

Question

Which is the greater quantity?

(a) The surface area of a regular tetrahedron with edges of length 1

(b) 2

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 :

$A = s^{2} \sqrt{3} =1^{2} \cdot \sqrt{3} = \sqrt{3}$

$\sqrt{3} \approx 1.7 < 2$ , so (b) is greater.

Compare your answer with the correct one above

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 :

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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