Calculating the volume of a tetrahedron - GMAT Quantitative

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Question

What is the volume of a right pyramid whose height is 20 and whose base is an equilateral triangle with sidelength 10?

Answer

The volume of a pyramid can be calculated using the formula

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

where is the height and is the area of the base.

Since the base is an equilateral triangle, its area can be calculated using the formula

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

Therefore, the volume can be rewritten as

$V = \frac{1}{3} \cdot \frac{s^{2} \sqrt{3}}{4} \cdot h$

Substitute :

$V = \frac{1}{3} \cdot \frac{10^{2}\cdot \sqrt{3}}{4} \cdot 20$

$= \frac{1}{3} \cdot \frac{100\cdot \sqrt{3}}{4} \cdot\frac{ 20}{1}$

$= \frac{1}{3} \cdot \frac{25\cdot \sqrt{3}}{1} \cdot\frac{ 20}{1}$

$=\frac{500 \sqrt{3}}{3}$

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Question

A right triangular pyramid has as its base an equilateral triangle with sidelength 10. Its height is 15.

Give its volume.

Answer

The base of the triangle has an area that can be found using the formula for the area of an equilateral triangle, substituting :

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

$B = \frac{10^{2}\sqrt{3}}{4}$

$B= \frac{100\sqrt{3}}{4}$

$B= 25\sqrt{3}$

Now, in the formula for the volume of a pyramid, substitute $B = 25\sqrt{3}, h = 15$ :

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

$V = \frac{1}{3}\cdot 25\sqrt{3} \cdot 15$

$V = 125\sqrt{3}$

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Question

The height of a right pyramid and the sidelength of its square base are equal. The perimeter of the base is one yard. Give its volume in cubic inches.

Answer

The perimeter of the base is one yard, or 36 inches; its sidelength - and, sunsequently, its height - are one-fourth of that, or 9 inches, and the area of the base is $9 ^{2} = 81$ square inches. The volume of a pyramid is one-third the product of its height and the area of its base, so substitute in the following:

$V = \frac{1}{3} Bh = \frac{1}{3} \cdot 81 \cdot 9 = 243$ cubic inches.

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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 60 units away from the origin on each of the three (mutually perpendicular) axes.

This is a triangular pyramid, and we can consider $\Delta OAB$ the (right triangular) base; its area is half the product of its legs, or

$B = \frac{1}{2} \cdot 60 \cdot 60 = 1,800$ .

The volume of the tetrahedron is one third the product of its base and its height, the latter of which is 60. Therefore,

$V = \frac{1}{3} \cdot 60 \cdot 1,800 = 36,000$ .

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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 is 60 and its height is 40, so its area is

$B = \frac{1}{2} \cdot 60 \cdot 40 = 1,200$

The fourth vertex is off the -plane; its perpendicular distance to the aforementioned face is its -coordinate, 20, so this is the height of the pyramid. The volume of the pyramid is

$V = \frac{1}{3} \cdot 1,200 \cdot 20 = 8,000$ .

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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

The tetrahedron looks like this:

is the origin and are the other three points, whose distances away from the origin on each of the three (perpendicular) axes are shown.

This is a triangular pyramid, and we can consider $\Delta OCB$ the (right triangular) base; its area is half the product of its legs, or

$B = \frac{1}{2} \cdot n \cdot 4n= 2n^{2}$ .

The volume of the tetrahedron is one third the product of its base and its height, the latter of which is . Therefore,

$V = \frac{1}{3} \cdot 9n \cdot 2n^{2} = 6 n^{3}$ .

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Question

A right pyramid has height $3^{x}$ ; its base is a square with four sides of length $3^{y}$ each. What is the volume of this pyramid?

Answer

The volume of a pyramid can be calculated using the formula

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

where is the area of the base and is the height. The base of the pyramid in question is a square; if we let its sidelength be , this base will be $s^{2}$ , and the volume formula will be

$V = \frac{1}{3} s ^{2}h$

Setting $s = 3^{y}$ and $h = 3^{x}$ :

$V =3 ^{-1} \cdot ( 3^{y}) ^{2} \cdot 3^{x}$

$V =3 ^{-1} \cdot 3^{2 y} \cdot 3^{x}$

$V =3 ^{-1+2 y+x}$

or

$V =3 ^{x+2y-1 }$ .

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Question

What is the volume of a right pyramid whose height is 20 and whose base is an equilateral triangle with sidelength 10?

Answer

The volume of a pyramid can be calculated using the formula

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

where is the height and is the area of the base.

Since the base is an equilateral triangle, its area can be calculated using the formula

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

Therefore, the volume can be rewritten as

$V = \frac{1}{3} \cdot \frac{s^{2} \sqrt{3}}{4} \cdot h$

Substitute :

$V = \frac{1}{3} \cdot \frac{10^{2}\cdot \sqrt{3}}{4} \cdot 20$

$= \frac{1}{3} \cdot \frac{100\cdot \sqrt{3}}{4} \cdot\frac{ 20}{1}$

$= \frac{1}{3} \cdot \frac{25\cdot \sqrt{3}}{1} \cdot\frac{ 20}{1}$

$=\frac{500 \sqrt{3}}{3}$

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Question

A right triangular pyramid has as its base an equilateral triangle with sidelength 10. Its height is 15.

Give its volume.

Answer

The base of the triangle has an area that can be found using the formula for the area of an equilateral triangle, substituting :

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

$B = \frac{10^{2}\sqrt{3}}{4}$

$B= \frac{100\sqrt{3}}{4}$

$B= 25\sqrt{3}$

Now, in the formula for the volume of a pyramid, substitute $B = 25\sqrt{3}, h = 15$ :

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

$V = \frac{1}{3}\cdot 25\sqrt{3} \cdot 15$

$V = 125\sqrt{3}$

Compare your answer with the correct one above

Question

The height of a right pyramid and the sidelength of its square base are equal. The perimeter of the base is one yard. Give its volume in cubic inches.

Answer

The perimeter of the base is one yard, or 36 inches; its sidelength - and, sunsequently, its height - are one-fourth of that, or 9 inches, and the area of the base is $9 ^{2} = 81$ square inches. The volume of a pyramid is one-third the product of its height and the area of its base, so substitute in the following:

$V = \frac{1}{3} Bh = \frac{1}{3} \cdot 81 \cdot 9 = 243$ cubic inches.

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 volume of this tetrahedron?

Answer

The tetrahedron looks like this:

is the origin and are the other three points, which are 60 units away from the origin on each of the three (mutually perpendicular) axes.

This is a triangular pyramid, and we can consider $\Delta OAB$ the (right triangular) base; its area is half the product of its legs, or

$B = \frac{1}{2} \cdot 60 \cdot 60 = 1,800$ .

The volume of the tetrahedron is one third the product of its base and its height, the latter of which is 60. Therefore,

$V = \frac{1}{3} \cdot 60 \cdot 1,800 = 36,000$ .

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 .

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 is 60 and its height is 40, so its area is

$B = \frac{1}{2} \cdot 60 \cdot 40 = 1,200$

The fourth vertex is off the -plane; its perpendicular distance to the aforementioned face is its -coordinate, 20, so this is the height of the pyramid. The volume of the pyramid is

$V = \frac{1}{3} \cdot 1,200 \cdot 20 = 8,000$ .

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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

The tetrahedron looks like this:

is the origin and are the other three points, whose distances away from the origin on each of the three (perpendicular) axes are shown.

This is a triangular pyramid, and we can consider $\Delta OCB$ the (right triangular) base; its area is half the product of its legs, or

$B = \frac{1}{2} \cdot n \cdot 4n= 2n^{2}$ .

The volume of the tetrahedron is one third the product of its base and its height, the latter of which is . Therefore,

$V = \frac{1}{3} \cdot 9n \cdot 2n^{2} = 6 n^{3}$ .

Compare your answer with the correct one above

Question

A right pyramid has height $3^{x}$ ; its base is a square with four sides of length $3^{y}$ each. What is the volume of this pyramid?

Answer

The volume of a pyramid can be calculated using the formula

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

where is the area of the base and is the height. The base of the pyramid in question is a square; if we let its sidelength be , this base will be $s^{2}$ , and the volume formula will be

$V = \frac{1}{3} s ^{2}h$

Setting $s = 3^{y}$ and $h = 3^{x}$ :

$V =3 ^{-1} \cdot ( 3^{y}) ^{2} \cdot 3^{x}$

$V =3 ^{-1} \cdot 3^{2 y} \cdot 3^{x}$

$V =3 ^{-1+2 y+x}$

or

$V =3 ^{x+2y-1 }$ .

Compare your answer with the correct one above

Question

What is the volume of a right pyramid whose height is 20 and whose base is an equilateral triangle with sidelength 10?

Answer

The volume of a pyramid can be calculated using the formula

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

where is the height and is the area of the base.

Since the base is an equilateral triangle, its area can be calculated using the formula

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

Therefore, the volume can be rewritten as

$V = \frac{1}{3} \cdot \frac{s^{2} \sqrt{3}}{4} \cdot h$

Substitute :

$V = \frac{1}{3} \cdot \frac{10^{2}\cdot \sqrt{3}}{4} \cdot 20$

$= \frac{1}{3} \cdot \frac{100\cdot \sqrt{3}}{4} \cdot\frac{ 20}{1}$

$= \frac{1}{3} \cdot \frac{25\cdot \sqrt{3}}{1} \cdot\frac{ 20}{1}$

$=\frac{500 \sqrt{3}}{3}$

Compare your answer with the correct one above

Question

A right triangular pyramid has as its base an equilateral triangle with sidelength 10. Its height is 15.

Give its volume.

Answer

The base of the triangle has an area that can be found using the formula for the area of an equilateral triangle, substituting :

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

$B = \frac{10^{2}\sqrt{3}}{4}$

$B= \frac{100\sqrt{3}}{4}$

$B= 25\sqrt{3}$

Now, in the formula for the volume of a pyramid, substitute $B = 25\sqrt{3}, h = 15$ :

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

$V = \frac{1}{3}\cdot 25\sqrt{3} \cdot 15$

$V = 125\sqrt{3}$

Compare your answer with the correct one above

Question

The height of a right pyramid and the sidelength of its square base are equal. The perimeter of the base is one yard. Give its volume in cubic inches.

Answer

The perimeter of the base is one yard, or 36 inches; its sidelength - and, sunsequently, its height - are one-fourth of that, or 9 inches, and the area of the base is $9 ^{2} = 81$ square inches. The volume of a pyramid is one-third the product of its height and the area of its base, so substitute in the following:

$V = \frac{1}{3} Bh = \frac{1}{3} \cdot 81 \cdot 9 = 243$ cubic inches.

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 volume of this tetrahedron?

Answer

The tetrahedron looks like this:

is the origin and are the other three points, which are 60 units away from the origin on each of the three (mutually perpendicular) axes.

This is a triangular pyramid, and we can consider $\Delta OAB$ the (right triangular) base; its area is half the product of its legs, or

$B = \frac{1}{2} \cdot 60 \cdot 60 = 1,800$ .

The volume of the tetrahedron is one third the product of its base and its height, the latter of which is 60. Therefore,

$V = \frac{1}{3} \cdot 60 \cdot 1,800 = 36,000$ .

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 .

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 is 60 and its height is 40, so its area is

$B = \frac{1}{2} \cdot 60 \cdot 40 = 1,200$

The fourth vertex is off the -plane; its perpendicular distance to the aforementioned face is its -coordinate, 20, so this is the height of the pyramid. The volume of the pyramid is

$V = \frac{1}{3} \cdot 1,200 \cdot 20 = 8,000$ .

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

,

where

Give its volume in terms of .

Answer

The tetrahedron looks like this:

is the origin and are the other three points, whose distances away from the origin on each of the three (perpendicular) axes are shown.

This is a triangular pyramid, and we can consider $\Delta OCB$ the (right triangular) base; its area is half the product of its legs, or

$B = \frac{1}{2} \cdot n \cdot 4n= 2n^{2}$ .

The volume of the tetrahedron is one third the product of its base and its height, the latter of which is . Therefore,

$V = \frac{1}{3} \cdot 9n \cdot 2n^{2} = 6 n^{3}$ .

Compare your answer with the correct one above

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