GMAT Quantitative - Calculating the volume of a tetrahedron

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.

$243 \textrm{ in}^{3}$

$729 \textrm{ in}^{3}$

$72 \textrm{ in}^{3}$

$216 \textrm{ in}^{3}$

$108 \textrm{ in}^{3}$

Explanation

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.

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?

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

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

$3 ^{x+4y-1 }$

$3 ^{x+2y+1 }$

$3 ^{2x+y+1 }$

Explanation

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 $h = 3^{x}$ :

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

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

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

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

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

Give its volume.

$125\sqrt{3}$

$375\sqrt{3}$

Explanation

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 $B = 25\sqrt{3}, h = 15$ :

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

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

$V = 125\sqrt{3}$

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

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

$500 \sqrt{3}$

$\frac{1.000 \sqrt{3}}{3}$

Explanation

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 :

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

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

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

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

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 .

$6 n^{3}$

$n^{3}$

$9n^{3}$

$3 n^{3}$

$\small 4 n^{3}$

Explanation

The tetrahedron looks like this:

Tetrahedron

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,

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

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?

Explanation

The tetrahedron looks like this:

Tetrahedron

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,

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

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates .

Give its volume.

Explanation

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:

Base

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

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

Calculating the volume of a tetrahedron

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