Tetrahedrons - PSAT Math

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Question

Refer to the above tetrahedron, or four-faced solid.The surface area of the tetrahedron is 444. Evaluate to the nearest tenth.

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Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength . Since the total surface area is 444, each triangle has area one fourth of this, or 111. To find , set in the formula for the area of an equilateral triangle:

$s\approx $\sqrt{256.34}$ \approx 16.0$

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Question

Note: Figure NOT drawn to scale.

The above triangular pyramid has volume 25. To the nearest tenth, evaluate .

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Answer

We are looking for the height of the pyramid.

The base is an equilateral triangle with sidelength 4, so its area can be calculated as follows:

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

$B = $$\frac{4^{2}$$ $\sqrt{3}$ }{4}$

$B = $\frac{16 \sqrt{3}$ }{4}$

$B = 4$\sqrt{3}$$

The height of a pyramid can be calculated using the fomula

$V = $\frac{1}{3}$Bh$

We set and $B = 4$\sqrt{3}$$ and solve for :

$$\frac{1}{3}$ \cdot 4$\sqrt{3}$ \cdot h = 25$

$h = $\frac{25 \cdot 3}{4 \sqrt{3}$} = $\frac{75}{4 \sqrt{3}$} \approx $\frac{75}{4 \cdot 1.7321}$ \approx 10.8$

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Question

Note: Figure NOT drawn to scale.

Give the volume (nearest tenth) of the above triangular pyramid.

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Answer

The height of the pyramid is . The base is an equilateral triangle with sidelength 4, so its area can be calculated as follows:

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

$B = $$\frac{4^{2}$$ $\sqrt{3}$ }{4}$

$B = $\frac{16 \sqrt{3}$ }{4}$

$B = 4$\sqrt{3}$$

The volume of a pyramid can be calculated using the fomula

$V = $\frac{1}{3}$Bh$

$V = $\frac{1}{3}$ \cdot 4$\sqrt{3}$ \cdot 8 = $\frac{32\sqrt{3}$ }{3} \approx $\frac{32 \cdot 1.7321}{3}$ \approx 18.5$

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Question

A regular tetrahedron has an edge length of . What is its volume?

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Answer

The volume of a tetrahedron is found with the equation , where represents the length of an edge of the tetrahedron.

Plug in 4 for the edge length and reduce as much as possible to find the answer:

$V=\frac{64}{6\sqrt{2}$}$

$V=\frac{32}{3\sqrt{2}$}$

The volume of the tetrahedron is $$\frac{32}{3\sqrt{2}$}$ .

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Question

Refer to the above tetrahedron, or four-faced solid.The surface area of the tetrahedron is 444. Evaluate to the nearest tenth.

Tap to reveal answer

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength . Since the total surface area is 444, each triangle has area one fourth of this, or 111. To find , set in the formula for the area of an equilateral triangle:

$s\approx $\sqrt{256.34}$ \approx 16.0$

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Question

Note: Figure NOT drawn to scale.

The above triangular pyramid has volume 25. To the nearest tenth, evaluate .

Tap to reveal answer

Answer

We are looking for the height of the pyramid.

The base is an equilateral triangle with sidelength 4, so its area can be calculated as follows:

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

$B = $$\frac{4^{2}$$ $\sqrt{3}$ }{4}$

$B = $\frac{16 \sqrt{3}$ }{4}$

$B = 4$\sqrt{3}$$

The height of a pyramid can be calculated using the fomula

$V = $\frac{1}{3}$Bh$

We set and $B = 4$\sqrt{3}$$ and solve for :

$$\frac{1}{3}$ \cdot 4$\sqrt{3}$ \cdot h = 25$

$h = $\frac{25 \cdot 3}{4 \sqrt{3}$} = $\frac{75}{4 \sqrt{3}$} \approx $\frac{75}{4 \cdot 1.7321}$ \approx 10.8$

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Question

Note: Figure NOT drawn to scale.

Give the volume (nearest tenth) of the above triangular pyramid.

Tap to reveal answer

Answer

The height of the pyramid is . The base is an equilateral triangle with sidelength 4, so its area can be calculated as follows:

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

$B = $$\frac{4^{2}$$ $\sqrt{3}$ }{4}$

$B = $\frac{16 \sqrt{3}$ }{4}$

$B = 4$\sqrt{3}$$

The volume of a pyramid can be calculated using the fomula

$V = $\frac{1}{3}$Bh$

$V = $\frac{1}{3}$ \cdot 4$\sqrt{3}$ \cdot 8 = $\frac{32\sqrt{3}$ }{3} \approx $\frac{32 \cdot 1.7321}{3}$ \approx 18.5$

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Question

A regular tetrahedron has an edge length of . What is its volume?

Tap to reveal answer

Answer

The volume of a tetrahedron is found with the equation , where represents the length of an edge of the tetrahedron.

Plug in 4 for the edge length and reduce as much as possible to find the answer:

$V=\frac{64}{6\sqrt{2}$}$

$V=\frac{32}{3\sqrt{2}$}$

The volume of the tetrahedron is $$\frac{32}{3\sqrt{2}$}$ .

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Question

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

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Answer

The area of an equilateral triangle is given by the formula

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

Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is

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

Substitute :

$SA = $s^{2}$$\sqrt{3}$ = $6^{2}$$\sqrt{3}$ = 36$\sqrt{3}$$ square inches.

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Question

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

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Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area

$A = $$\frac{s^{2}$$ $\sqrt{3}$ }{4} = $$\frac{7^{2}$$ $\sqrt{3}$ }{4} = $\frac{49 \sqrt{3}$ }{4} \approx $\frac{49 \cdot 1.7321}{4}$ \approx 21.22$

The total surface area is four times this, or about $4 \cdot 21.22 = 84.88$ .

Rounded, this is 84.9.

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Question

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

Tap to reveal answer

Answer

The area of an equilateral triangle is given by the formula

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

Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is

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

Substitute :

$SA = $s^{2}$$\sqrt{3}$ = $6^{2}$$\sqrt{3}$ = 36$\sqrt{3}$$ square inches.

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Question

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

Tap to reveal answer

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area

$A = $$\frac{s^{2}$$ $\sqrt{3}$ }{4} = $$\frac{7^{2}$$ $\sqrt{3}$ }{4} = $\frac{49 \sqrt{3}$ }{4} \approx $\frac{49 \cdot 1.7321}{4}$ \approx 21.22$

The total surface area is four times this, or about $4 \cdot 21.22 = 84.88$ .

Rounded, this is 84.9.

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Question

Note: Figure NOT drawn to scale.

The above triangular pyramid has volume 25. To the nearest tenth, evaluate .

Tap to reveal answer

Answer

We are looking for the height of the pyramid.

The base is an equilateral triangle with sidelength 4, so its area can be calculated as follows:

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

$B = $$\frac{4^{2}$$ $\sqrt{3}$ }{4}$

$B = $\frac{16 \sqrt{3}$ }{4}$

$B = 4$\sqrt{3}$$

The height of a pyramid can be calculated using the fomula

$V = $\frac{1}{3}$Bh$

We set and $B = 4$\sqrt{3}$$ and solve for :

$$\frac{1}{3}$ \cdot 4$\sqrt{3}$ \cdot h = 25$

$h = $\frac{25 \cdot 3}{4 \sqrt{3}$} = $\frac{75}{4 \sqrt{3}$} \approx $\frac{75}{4 \cdot 1.7321}$ \approx 10.8$

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Question

Note: Figure NOT drawn to scale.

Give the volume (nearest tenth) of the above triangular pyramid.

Tap to reveal answer

Answer

The height of the pyramid is . The base is an equilateral triangle with sidelength 4, so its area can be calculated as follows:

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

$B = $$\frac{4^{2}$$ $\sqrt{3}$ }{4}$

$B = $\frac{16 \sqrt{3}$ }{4}$

$B = 4$\sqrt{3}$$

The volume of a pyramid can be calculated using the fomula

$V = $\frac{1}{3}$Bh$

$V = $\frac{1}{3}$ \cdot 4$\sqrt{3}$ \cdot 8 = $\frac{32\sqrt{3}$ }{3} \approx $\frac{32 \cdot 1.7321}{3}$ \approx 18.5$

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Question

A regular tetrahedron has an edge length of . What is its volume?

Tap to reveal answer

Answer

The volume of a tetrahedron is found with the equation , where represents the length of an edge of the tetrahedron.

Plug in 4 for the edge length and reduce as much as possible to find the answer:

$V=\frac{64}{6\sqrt{2}$}$

$V=\frac{32}{3\sqrt{2}$}$

The volume of the tetrahedron is $$\frac{32}{3\sqrt{2}$}$ .

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Question

Refer to the above tetrahedron, or four-faced solid.The surface area of the tetrahedron is 444. Evaluate to the nearest tenth.

Tap to reveal answer

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength . Since the total surface area is 444, each triangle has area one fourth of this, or 111. To find , set in the formula for the area of an equilateral triangle:

$s\approx $\sqrt{256.34}$ \approx 16.0$

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Question

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

Tap to reveal answer

Answer

The area of an equilateral triangle is given by the formula

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

Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is

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

Substitute :

$SA = $s^{2}$$\sqrt{3}$ = $6^{2}$$\sqrt{3}$ = 36$\sqrt{3}$$ square inches.

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Question

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

Tap to reveal answer

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area

$A = $$\frac{s^{2}$$ $\sqrt{3}$ }{4} = $$\frac{7^{2}$$ $\sqrt{3}$ }{4} = $\frac{49 \sqrt{3}$ }{4} \approx $\frac{49 \cdot 1.7321}{4}$ \approx 21.22$

The total surface area is four times this, or about $4 \cdot 21.22 = 84.88$ .

Rounded, this is 84.9.

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Question

Refer to the above tetrahedron, or four-faced solid.The surface area of the tetrahedron is 444. Evaluate to the nearest tenth.

Tap to reveal answer

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength . Since the total surface area is 444, each triangle has area one fourth of this, or 111. To find , set in the formula for the area of an equilateral triangle:

$s\approx $\sqrt{256.34}$ \approx 16.0$

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Question

Note: Figure NOT drawn to scale.

The above triangular pyramid has volume 25. To the nearest tenth, evaluate .

Tap to reveal answer

Answer

We are looking for the height of the pyramid.

The base is an equilateral triangle with sidelength 4, so its area can be calculated as follows:

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

$B = $$\frac{4^{2}$$ $\sqrt{3}$ }{4}$

$B = $\frac{16 \sqrt{3}$ }{4}$

$B = 4$\sqrt{3}$$

The height of a pyramid can be calculated using the fomula

$V = $\frac{1}{3}$Bh$

We set and $B = 4$\sqrt{3}$$ and solve for :

$$\frac{1}{3}$ \cdot 4$\sqrt{3}$ \cdot h = 25$

$h = $\frac{25 \cdot 3}{4 \sqrt{3}$} = $\frac{75}{4 \sqrt{3}$} \approx $\frac{75}{4 \cdot 1.7321}$ \approx 10.8$

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