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Advanced Geometry : Solid Geometry

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Example Question #51 : Advanced Geometry

A regular tetrahedron has surface area 1,000. Which of the following comes closest to the length of one edge?

Possible Answers:

Correct answer:

Explanation:

A regular tetrahedron has six congruent edges and, as its faces, four congruent equilateral triangles. If we let be the length of one edge, each face has as its area

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

the total surface area of the tetrahedron is therefore four times this, or

S = 4A

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

$S = s^{2} \sqrt{3}$

Set S = 1,000 and solve for :

$s^{2} \sqrt{3} = 1,000$

Divide by $\sqrt{3}$ :

$\frac{s^{2} \sqrt{3} }{ \sqrt{3} }= \frac{1,000}{ \sqrt{3} }$

$s^{2} = \frac{1,000}{1.7321 }$

$s^{2} \approx 577.3503$

Take the square root of both sides:

$s \approx \sqrt{577.3503}$

$s \approx 24.0281$

Of the given choices, 20 comes closest.

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Example Question #52 : Advanced Geometry

Tetra

The above figure shows a triangular pyramid, or tetrahedron, on the three-dimensional coordinate axes. The tetrahedron has volume 1,000. Which of the following is closest to the value of ?

Possible Answers:

Correct answer:

Explanation:

If we take the triangle on the -plane to be the base of the pyramid, this base has legs both of length ; its area is half the product of the lengths which is

$B = \frac{1}{2} ab = \frac{1}{2} r ^{2}$

Its height is the length of the side along the -axis, which is also of length .

The volume of a pyramid is equal to one third the product of its height and the area of its base, so

$V = \frac{1}{3} \cdot r \cdot \frac{1}{2} r ^{2}$

$= \frac{1}{3} \cdot \frac{1}{2}\cdot r \cdot r ^{2}$

$= \frac{1}{6} r ^{3}$

Setting the volume equal to 1,000, we can solve for :

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

Multiply both sides by 6:

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

$r ^{3} =6 ,000$

Take the cube root of both sides:

$r =\sqrt[3]{6 ,000}$

$r \approx 18.2$

The closest choice is 20.

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Example Question #51 : Advanced Geometry

Given a regular tetrahedron with an edge of $\small 10 cm$ , what is the height (or diagonal)? The height is the line drawn from one vertex perpendicular to the opposite face. Tetrahedron_10

Possible Answers:

None of the above.

$\small \small \frac{1}{5}\sqrt{3} cm$

$\small \small \frac{\sqrt{2}}{\sqrt{3}} cm$

$\small \small 10\sqrt{\frac{2}{3}} cm$

$\small \small 5\sqrt{3} cm$

Correct answer:

$\small \small 10\sqrt{\frac{2}{3}} cm$

Explanation:

The height of a regular tetrahedron can be derived from the formula

$\small h= \sqrt{\frac{2}{3}}a$ where $\small a$ is the length of one edge.

Therefore, plugging in the side length of 10cm ,

$h=\sqrt\frac{2}{3}\cdot 10$ .

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Example Question #1 : How To Find The Diagonal Of A Tetrahedron

Given a regular tetrahedron with an edge of $\small \small 20 cm$ , what is the height (or diagonal)? The height is the line drawn from one vertex perpendicular to the opposite face.

Tetrahedron_20

Possible Answers:

$\small 10\sqrt{2}cm$

$\small 10\sqrt{\frac{2}{3}}cm$

$\small 20\sqrt{2}cm$

None of the above.

$\small 20\sqrt{\frac{2}{3}}cm$

Correct answer:

$\small 20\sqrt{\frac{2}{3}}cm$

Explanation:

The height of a regular tetrahedron can be derived from the formula $\small h= \sqrt{\frac{2}{3}}a$ where $\small a$ is the length of one edge.

Plugging in a=20 we can solve for .

$\small h= \sqrt{\frac{2}{3}}\cdot 20$

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Example Question #1 : How To Find The Diagonal Of A Tetrahedron

Find the height of this regular tetrahedron:

Tetrahedron

Possible Answers:

$2\sqrt{2}$

$\sqrt{2}$

Correct answer:

$2\sqrt{2}$

Explanation:

The height of a regular tetrahedron can be found using the formula $h=\sqrt{\frac{2}{3}}\cdot s$ where s is the length of the sides.

In this case, the sides have length $2\sqrt{3}$ , so we are multiplying $\sqrt{\frac{2}{3}}\cdot 2\sqrt{3}$ .

We can simplify this by multiplying the numbers inside the radical:

$2\sqrt{\frac{2}{3}\cdot 3}$ , which simplifies to $2\sqrt{2}$ .

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Example Question #1 : How To Find The Surface Area Of A Tetrahedron

What is the surface area of the following tetrahedron? Assume the figure is a regular tetrahedron.

Tetrahedron

Possible Answers:

$8\sqrt{3}m^2$

$\frac{\sqrt{3}}{2}m^2$

$3\sqrt{3}m^2$

$2\sqrt{3}m^2$

$4\sqrt{3}m^2$

Correct answer:

$4\sqrt{3}m^2$

Explanation:

A tetrahedron is a three-dimensonal figure where each side is an equilateral triangle. Therefore, each angle in the triangle is $60^{\circ}$ .

In the figure, we know the value of the side $(2\: m)$ and the value of the base $(1\: m)$ . Since dividing the triangle by half creates a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, we know the value of must be $\sqrt{3}\: m$ .

Therefore, the area of one side of the tetrahedron is:

$A = (b)(h) = (1\: m)(\sqrt{3}\: m) = \sqrt{3}\: m^2$

Since there are four sides of a tetrahedron, the surface area is:

$SA = 4 (b)(h) = 4\sqrt{3}\: m^2$

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Example Question #1 : How To Find The Surface Area Of A Tetrahedron

A regular tetrahedron has side lengths s=1 . What is the surface area of the described solid?

Possible Answers:

$4\sqrt{3}$

$2\sqrt3$

$\sqrt{3}$

Correct answer:

$\sqrt{3}$

Explanation:

The area of one face of the triangle can be found either through trigonometry or the Pythagorean Theorem.

Since all the sides of the triangle are , the height is then $\frac{\sqrt{3}}{2}$ , so the area of each face is:

$\frac{bh}{2}=\frac{1}{2} \cdot 1 \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}$

There are four faces, so the area of the tetrahedron is:

$4 \cdot \frac{bh}{2}=\sqrt3$

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Example Question #1 : How To Find The Surface Area Of A Tetrahedron

Find the surface area of a regular tetrahedron with a side length of .

Possible Answers:

$2\sqrt3$

$4\sqrt3$

$16\sqrt3$

$8\sqrt3$

Correct answer:

$16\sqrt3$

Explanation:

Use the following formula to find the surface area of a regular tetrahedron.

$\text{Surface Area}=\sqrt3 (side^2)$

Now, substitute in the value of the side length into the equation.

$\text{Surface Area}=\sqrt3(4^2)=16\sqrt3$

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Example Question #51 : Advanced Geometry

Find the surface area of a regular tetrahedron with a side length of .

Possible Answers:

$12\sqrt3$

$36\sqrt3$

$144\sqrt3$

$96\sqrt3$

Correct answer:

$144\sqrt3$

Explanation:

Use the following formula to find the surface area of a regular tetrahedron.

$\text{Surface Area}=\sqrt3 (side^2)$

Now, substitute in the value of the side length into the equation.

$\text{Surface Area}=\sqrt3(12^2)=144\sqrt3$

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Example Question #1 : How To Find The Surface Area Of A Tetrahedron

In terms of , find the surface area of a regular tetrahedron that has a side length of .

Possible Answers:

$25x\sqrt3$

$5x^2\sqrt3$

$25x^2\sqrt3$

$125x^2\sqrt3$

Correct answer:

$25x^2\sqrt3$

Explanation:

Use the following formula to find the surface area of a regular tetrahedron.

$\text{Surface Area}=\sqrt3 (side^2)$

Now, substitute in the value of the side length into the equation.

$\text{Surface Area}=\sqrt3(5x)^2=25x^2\sqrt3$

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Advanced Geometry : Solid Geometry

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #51 : Advanced Geometry

Example Question #52 : Advanced Geometry

Example Question #51 : Advanced Geometry

Example Question #1 : How To Find The Diagonal Of A Tetrahedron

Example Question #1 : How To Find The Diagonal Of A Tetrahedron

Example Question #1 : How To Find The Surface Area Of A Tetrahedron

Example Question #1 : How To Find The Surface Area Of A Tetrahedron

Example Question #1 : How To Find The Surface Area Of A Tetrahedron

Example Question #51 : Advanced Geometry

Example Question #1 : How To Find The Surface Area Of A Tetrahedron

All Advanced Geometry Resources

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