Tetrahedrons

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GMAT Quantitative › Tetrahedrons

Questions 1 - 10

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.

The slant height of a pyramid is one and one-half times the perimeter of its square base. The base has sides of length 15 inches. What is the surface area of the pyramid?

$2,925 \textrm{ in} ^{2}$

$2,700 \textrm{ in} ^{2}$

$2,025 \textrm{ in} ^{2}$

$1,800 \textrm{ in} ^{2}$

$4,950 \textrm{ in} ^{2}$

Explanation

The square base of the pyramid has four sides with length 15 inches, making its perimeter four times that, or 60 inches. The slant height is

$1 \frac{1}{2} \times 60 = 90$ inches.

Therefore, the area of the base is $B = 15 ^{2} = 225$ square inches.

Each of the four lateral triangles has area $\frac{1}{2} \cdot 15 \cdot 90 = 675$ square inches.

The total surface area is $225 + 4 \cdot 675 = 2,925$ square 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 $h = 3^{x}$ :

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

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

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

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

Tetra_1

Refer to the above diagram, which shows a tetrahedron.

, and . Give the surface area of the tetrahedron.

$675 + 225 \sqrt{3}$

$900 \sqrt{3}$

$1,350 + 225 \sqrt{3}$

Explanation

Three of the surfaces of the tetrahedron - $\bigtriangleup ADB$ , $\bigtriangleup ADC$ , and $\bigtriangleup BDC$ - are isosceles right triangles with hypotenuse 30, so by the 45-45-90 Theorem, each leg measures this length divided by $\sqrt{2}$ , or

$\frac{30 }{\sqrt{2}}= \frac{30 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}= \frac{30 \cdot \sqrt{2}}{2} = 15 \sqrt{2}$ .

The area of each of these triangles is half the product of its legs, so each area is

$\frac{1}{2}\cdot 15 \sqrt{2} \cdot 15 \sqrt{2} = \frac{1}{2}\cdot 15 \cdot 15\cdot \sqrt{2} \cdot \sqrt{2}= \frac{1}{2}\cdot 15 \cdot 15\cdot 2=225$

Also, the legs are of the same measure among the triangles, the hypotenuses are as well, so the fourth surface $\bigtriangleup ABC$ is an equilateral triangle. Its sidelength is , so we use the equilateral triangle area formula to calculate its area:

$\frac{30 ^{2}\cdot \sqrt{3}}{4} = \frac{900\cdot \sqrt{3}}{4} = 225 \sqrt{3}$

Add the areas of the faces:

$225 + 225 + 225+ 225 \sqrt{3} = 675 + 225 \sqrt{3}$

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 :

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

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

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

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

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

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

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

$V = 125\sqrt{3}$

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,

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

A regular tetrahedron comprises four faces, each of which is an equilateral triangle. If the sum of the lengths of its edges is 120, what is its surface area?

$400 \sqrt{3}$

$900 \sqrt{3}$

$225\sqrt{3}$

$600\sqrt{3}$

$120\sqrt{3}$

Explanation

As shown in the diagram below, a regular tetrahedron has six congruent edges, so each has length $\small 120 \div 6 = 20$ :

Tetra_2

The area of one face is the area of an equilateral triangle with sidelength 20, which is

$\small \frac{20^{2}\sqrt{3}}{4} = \frac{400\sqrt{3}}{4} = 100 \sqrt{3}$

The total surface area is four times this, or $\small 400 \sqrt{3}$ .

In three-dimensional space, the four vertices of a tetrahedron - a solid with four faces - have Cartesian coordinates $\small (0,0,0), (60,0,0), (0,60,0), (60,0,0)$ .

Give the surface area of the tetrahedron.

$\small 5,400+ 1,800 \sqrt{3}$

$\small \small 5,400+ 1,800 \sqrt{2}$

$\small 7,200 \sqrt{3}$

$\small \small 7,200 \sqrt{2}$

$\small 7,200$

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 (perpendicular) axes.

The bottom, front, and left faces are each right triangles whose legs each measure 60. Each face has area

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

The remaining face has three edges each a hypotenuse of one of three congruent right triangles, so its sides are congruent, and it is an equilateral triangle. Its sidelength can be found via the 45-45-90 Theorem to be $\small 60 \sqrt{ 2}$ , so its area is

$\small \frac{\left ( 60 \sqrt{2}\right )^{2} \cdot \sqrt{3}}{4} = \frac{3,600\left ( \sqrt{2}\right )^{2} \cdot \sqrt{3}}{4} = \frac{3,600 \cdot 2 \cdot \sqrt{3}}{4}=1,800 \sqrt{3}$