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

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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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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,

$V = \frac{1}{3} \cdot 12 \cdot 72 = 288$ .

Compare your answer with the correct one above

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:

$V = \frac{1}{3} \cdot 30 \cdot 300 \sqrt{3} = 3,000\sqrt{3}$

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 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

$V = \frac{1}{3} \cdot 90 \cdot 6 = 180$

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 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

$V = \frac{1}{3} \cdot 162 \cdot 9 = 486$

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 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,

$V = \frac{1}{3} \cdot 12 \cdot 72 = 288$ .

Compare your answer with the correct one above

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:

$V = \frac{1}{3} \cdot 30 \cdot 300 \sqrt{3} = 3,000\sqrt{3}$

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 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

$V = \frac{1}{3} \cdot 90 \cdot 6 = 180$

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 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

$V = \frac{1}{3} \cdot 162 \cdot 9 = 486$

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

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

$V = \frac{1}{3} \cdot 90 \cdot 2N = 60N$

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 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,

$V = \frac{1}{3} \cdot 12 \cdot 72 = 288$ .

Compare your answer with the correct one above

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:

$V = \frac{1}{3} \cdot 30 \cdot 300 \sqrt{3} = 3,000\sqrt{3}$

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 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

$V = \frac{1}{3} \cdot 90 \cdot 6 = 180$

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 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

$V = \frac{1}{3} \cdot 162 \cdot 9 = 486$

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

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

$V = \frac{1}{3} \cdot 90 \cdot 2N = 60N$

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

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

$V = \frac{1}{3} \cdot 90 \cdot 2N = 60N$

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 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,

$V = \frac{1}{3} \cdot 12 \cdot 72 = 288$ .

Compare your answer with the correct one above

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:

$V = \frac{1}{3} \cdot 30 \cdot 300 \sqrt{3} = 3,000\sqrt{3}$

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 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

$V = \frac{1}{3} \cdot 90 \cdot 6 = 180$

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 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

$V = \frac{1}{3} \cdot 162 \cdot 9 = 486$

Compare your answer with the correct one above

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