Tetrahedrons - GMAT Quantitative

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Question

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.

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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 (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

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

The total area is

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

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Question

The volume of a tetrahedron is . What is the edge of the tetrahedron?

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Answer

Write the formula to solve for the edge of a tetrahedron.

$e=\sqrt2 \sqrt[3]{3V}$

Substitute the given volume to the equation and solve.

$e=\sqrt2 \sqrt[3]{3(9)}= \sqrt2 \sqrt[3]{27}=3\sqrt2$

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Question

Suppose the volume of the tetrahedron is $\frac{1}{3}$ . What is the edge of the tetrahedron?

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Answer

Write the formula to solve for the edge of a tetrahedron.

$e=\sqrt2 \sqrt[3]{3V}$

Substitute the volume and solve.

$e=\sqrt2 \sqrt[3]{3(\frac{1}{3})}= \sqrt2 \sqrt[3]{1}=\sqrt2$

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Question

Find the exact edge length of a tetrahedron if the volume is .

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Answer

Write the tetrahedron formula to solve for the edge.

$e=\sqrt2 \sqrt[3]{3V}$

Substitute the volume and simplify.

$e=\sqrt2 \sqrt[3]{3(3)}= \sqrt2 \sqrt[3]{9}$

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Question

The height of one of the equilateral triangle faces on a tetrahedron is $6\sqrt{3}$ . What is the side length of the tetrahedron?

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Answer

Because a tetrahedron has four congruent equilateral triangles as its faces, we know the three equal angles of each face, as with all equilateral triangles, are each $60^{\circ}$ . We are given the height of a face, which is the length of a line that bisects one of the $60^{\circ}$ angles and forms a right triangle with the side length as its hypotenuse. This means the angle between the height and the side length is $30^{\circ}$ , whose cosine is equal to the adjacent side, the height, over the hypotenuse, the side length. This gives us:

$\cos 30^{\circ}=\frac{h}{a}$

$\frac{\sqrt{3}}{2}=\frac{6\sqrt{3}}{a}$

$a=\frac{6\sqrt{3}}{\frac{\sqrt{3}}{2}}=6\sqrt{3}\cdot \frac{2}{\sqrt{3}}=12$

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Question

A certain tetrahedron has a surface area of $16\sqrt{3}$ . What is the length of an edge of the tetrahedron?

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Answer

A tetrahedron has four congruent equilateral triangle faces, so its surface area is four times the area of one of these equilateral triangles. This gives us:

$SA=4\left(\frac{1}{2}bh\right)=2bh$

Where is the base of the triangle, or the edge length of the tetrahedron, and is the height of each triangular face. We want to solve for , but we are only given the surface area and not the height, so we need to express this value in terms of the base. One way of doing this is by recognizing that the height of each equilateral triangle face bisects an angle and forms two congruent right triangles for which it is the base and the edge length, or , is the hypotenuse. Each angle in an equilateral triangle is $60^{\circ}$ , so if one is bisected by the height then the angle between it and the hypotenuse is $30^{\circ}$ . The cosine of this angle is equal to the height over the hypotenuse, which gives us:

$\cos30^{\circ}=\frac{h}{b}\rightarrow h=b\cos30^{\circ}\rightarrow h=\frac{\sqrt{3}}{2}b$

Now that we have an expression for the height in terms of the base, we can plug this value into the equation for surface area and be left with only one unknown, the base that is equivalent to the edge length for which we want to solve:

$SA=2bh=2b\left(\frac{\sqrt{3}}{2}b\right)=\sqrt{3}b^2$

$16\sqrt{3}=\sqrt{3}b^2$

$b^2=16\rightarrow b=4$

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Question

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?

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Answer

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.

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Question

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?

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Answer

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

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 $\small 400 \sqrt{3}$ .

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Question

Refer to the above diagram, which shows a tetrahedron.

, and . Give the surface area of the tetrahedron.

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Answer

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:

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

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Question

A regular tetrahedron is a solid with four faces, each of which is an equilateral triangle.

If the lengths of all of the edges of a regular tetrahedron are added, the total length is 120. What is the surface area of the tetrahedron?

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Answer

A tetrahedron looks like this:

The tetrahedron has six edges, and in a regular tetrahedron, they are congruent, so each edge has length $120 \div 6 = 20$ .

The area of one face of the tetrahedron, it being an equilateral triangle, can be calulated using the formula

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

There are four congruent faces, so the total surface area is

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

Since , the surface area of the tetrahedron is

$S= 20^{2}\cdot \sqrt{3} =400 \sqrt{3}$

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Question

A regular tetrahedron is a solid with four faces, each of which is an equilateral triangle.

Each edge of a regular tetrahedron has length $6 \sqrt{3}$ . What is the surface area of the tetrahedron?

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Answer

The area of one face of the tetrahedron, it being an equilateral triangle, can be calulated using the formula

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

There are four congruent faces, so the total surface area is

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

Since $s = 6 \sqrt{3}$ , the surface area of the tetrahedron is

$S= \left ( 6 \sqrt{3}\right )^{2}\cdot \sqrt{3} =6^{2} \cdot \left ( \sqrt{3}\right )^{2}\cdot \sqrt{3} = 36 \cdot 3 \cdot \sqrt{3} = 108 \sqrt{3}$

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Question

Evaluate the surface area of the above tetrahedron.

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Answer

Three of the faces of the tetrahedron are isosceles right triangles with legs of length 8, and, subsequently, by the 45-45-90 Theorem, hypotenuses of length $8 \sqrt{2}$ . The fourth face, consequently, is an equilateral triangle with three sides of length $8 \sqrt{2}$ .

Each of the three right triangle faces has area equal to half the product of its legs , which is

$\frac{1}{2} \cdot 8 \cdot 8 = 32$ .

The equilateral face has as its area

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

$= \frac{ (8 \sqrt{2})^{2}\sqrt{3}}{4}$

$= \frac{ 8^{2}\cdot ( \sqrt{2})^{2}\cdot \sqrt{3}}{4}$

$= \frac{ 64\cdot 2\cdot \sqrt{3}}{4}$

$= \frac{ 128\cdot \sqrt{3}}{4}$

$= 32 \sqrt{3}$

The sum of the areas of the faces is

$32 \sqrt{3} + 32+32+32 = 96+ 32 \sqrt{3}$

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Question

The cube in the above figure has surface area 384. Give the surface area of the tetrahedron with vertices , shown in red.

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Answer

The surface area formula can be used to find the length of each edge of the cube:

$6s^{2}=384$

$s^{2}= 64$

Three faces of the tetrahedron - $\bigtriangleup DHE$ , $\bigtriangleup DHG$ , $\bigtriangleup EHG$ - are right triangles with legs of length 8, so the area of each is half the product of the lengths of their legs:

$\frac{1}{2} \cdot 8 \cdot 8 =32$ .

Each triangle is isosceles, so, by the 45-45-90 Theorem, each of their hypotentuses measures $\sqrt{2}$ times a leg, or $8\sqrt{2}$ . is therefore an equilateral triangle with sidelenghth $8\sqrt{2}$ . Its area can be found as follows:

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

$= \frac{(8\sqrt{2})^{2}\sqrt{3}}{4}$

$= \frac{8^{2}\cdot (\sqrt{2})^{2} \cdot \sqrt{3}}{4}$

$= \frac{64\cdot 2 \cdot \sqrt{3}}{4}$

$= 32 \sqrt{3}$

The total surface area is

$32+32+32+ 32 \sqrt{3} = 96 + 32 \sqrt{3}$

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Question

The above diagram shows a regular right triangular pyramid. Its base $\bigtriangleup ABC$ is an equilateral triangle; the other three faces are congruent isosceles triangles, with $\overline{VM}$ an altitude of $\bigtriangleup AVC$ . Give the surface area of the pyramid.

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Answer

The base is an equilateral triangle with sidelength 12, so its area can be calculated as follows:

$\frac{12^{2}\cdot \sqrt{3}}{4} = \frac{144 \sqrt{3}}{4} = 36 \sqrt{3}$ .

Each of the three other faces is congruent, with base 12. The area of each is the product of its base and its height. To find the common height, we examine $\bigtriangleup VMC$ , which, since $\overline{VM}$ is an altitude of isosceles $\bigtriangleup AVC$ , is a right triangle with hypotenuse of length 18 and one leg of length $MC = \frac{1}{2} AC = \frac{1}{2} \cdot 12 = 6$ . We can find using the Pythagorean Theorem:

$VM = \sqrt{\left (VC \right )^{2} - \left ( CM\right )^{2}}$

$= \sqrt{18^{2} - 6^{2}}$

$= \sqrt{288}$

$= \sqrt{144} \cdot \sqrt{2}$

$=12\sqrt{2}$

The area of $\bigtriangleup AVC$ is half the product of this height and the base:

$\frac{1}{2} \cdot 12 \cdot 12 \sqrt{2} = 72 \sqrt{2}$

All three lateral faces have this area.

Now add the areas of the four faces:

$36 \sqrt{3}+72 \sqrt{2}+72 \sqrt{2}+72 \sqrt{2} = 36 \sqrt{3}+216 \sqrt{2}$

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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?

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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.

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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.

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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?

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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.

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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 .

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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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