Tetrahedrons - GMAT Quantitative

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Question

Find the length of the edge of a tetrahedron.

Statement 1: The volume is 6.

Statement 2: The surface area is 6.

Answer

Statement 1:) The volume is 6.

Write the formula to find the edge of the tetrahedron given the volume.

$V=\frac{e^3}{6\sqrt2}$

Given the volume, it is possible to find the edge of the tetrahedron.

Statement 2:) The surface area is 6.

Write the formula to find the edge of the tetrahedron given the surface.

$SA=\sqrt3 e^2$

Substitute the surface area to find the edge.

Therefore:

$\textup{EACH statement ALONE is sufficient.}$

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Question

Note: Figure NOT drawn to scale.

Refer to the above tetrahedron or triangular pyramid. .

Calculate the surface area of the tetrahedron.

Statement 1: $\bigtriangleup ABC$ has perimeter 60.

Statement 2: $\bigtriangleup ADC$ has area 100.

Answer

$\bigtriangleup ADC$ , $\bigtriangleup ADB$ , and $\bigtriangleup BDC$ , all being right triangles with the same leg lengths, are congruent, and, consequently, their diagonals are congruent, making $\bigtriangleup ABC$ equilateral.

Assume Statement 1 alone. The sidelength of equilateral $\bigtriangleup ABC$ is one third of its perimeter of 60, or 20. This is also the common length of the hypotenuses of isoseles right triangles $\bigtriangleup ADC$ , $\bigtriangleup ADB$ , and $\bigtriangleup BDC$ , so by the 45-45-90 Theorem, each leg length can be computed by dividing 20 by $\sqrt{2}$ :

$\frac{20}{\sqrt{2}} = \frac{20\cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}} = \frac{20 \sqrt{2}}{2} = 10 \sqrt{2}$

Assume Statement 2 alone. $\bigtriangleup ADC$ has area 100; since the area of a right triangle is half the product of the lengths of its legs, we can find the common leg length using this formula:

$\frac{1}{2} s^{2} = A$

$\frac{1}{2} s^{2} = 100$

$s^{2} = 200$

$s =\sqrt{200} = \sqrt{100}\cdot \sqrt{2}= 10 \sqrt{2}$

This is the common length of the legs of the three right triangles; by the 45-45-90 Theorem, each hypotenuse - and each side of equilateral $\bigtriangleup ADC$ - is $\sqrt{2}$ times this, or $10 \sqrt{2} \cdot \sqrt{2} = 10 \cdot 2 = 20$ .

Therefore, from either statement alone, the lengths of all sides of the tetrahedron can be found, and the area formulas for a right triangle and an equilateral can be applied to find the areas of all four faces.

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Question

What is the surface area of the tetrahedron?

The length of an edge measures $3\sqrt{2}cm$ .

The volume of the tetrahedron is .

Answer

The surface area of a tetrahedron is found by $SA=a^2 \sqrt{3}$ where represents the edge value.

Situation 1: We're given our value so we just need to plug it into our equation.

$SA=(3\sqrt{2})^2 \sqrt{3}$

$SA=18\sqrt{3}$

Situation 2: We use the given volume to solve for the length of the edge.

$V=\frac{a^3}{6\sqrt{2}}=9$

$a=3\sqrt{2}$

Now that we have a length, we can plug it into the equation for the surface area:

$SA=(3\sqrt{2})^2 \sqrt{3}=18\sqrt{2}$

Thus, each statement alone is sufficient to answer the question.

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Question

In the above diagram, a tetrahedron—a triangular pyramid—with vertices is shown inside a cube. Give the volume of the tetrahedron.

Statement 1: The cube can be inscribed inside a sphere with volume $972 \pi$ .

Statement 2: A sphere with surface area $108 \pi$ can be inscribed inside the cube.

Answer

The volume of the pyramid is one third the product of height and the area of its base, which in turn, since here it is a right triangle, is half the product of the lengths and of its legs. Since the prism in the figure is a cube, the three lengths are equal, so we can set each to . The volume of the pyramid is

$V = \frac{1}{3} \cdot s\cdot \frac{1}{2} s\cdot s$

$V = \frac{1}{6} s^{3}$

Therefore, knowing the length of one edge of the cube is sufficient to determine the volume of the pyramid.

Assume Statement 1 alone. If the volume of the circumscribing sphere is known to be $972 \pi$ , the radius can be calculated as follows:

$\frac{4}{3} \pi r^{3} = V$

$\frac{4}{3} \pi r^{3} =972 \pi$

$\frac{3}{4\pi} \cdot \frac{4}{3} \pi r^{3} =\frac{3}{4\pi} \cdot 972 \pi$

$r^{3} =729$

$r = \sqrt[3]{729}= 9$

The diameter, which is twice this, or 18, is the length of a diagonal of the cube. By the three-dimensional extension of the Pythagorean Theorem, the relationship of this length to the side length of the cube is

$s^{2}+s^{2}+s^{2} = d^{2}$ , or

$3s^{2} = d^{2}$

so

$3s^{2} = 18^{2}$

$s^{2} = \frac{18^{2}}{3} = 108$

$s = \sqrt{108} = 6 \sqrt{3}$ ,

Assume Statement 2 alone. If the surface area of the inscribed sphere is known to be $108 \pi$ , then its radius can be calculated as follows:

$4 \pi r^{2} = A$

$4 \pi r^{2} = 108 \pi$

$\frac{4 \pi r^{2} }{4 \pi }=\frac{ 108 \pi}{4 \pi }$

$r^{2} = 27$

$r = \sqrt{27} = 3\sqrt{3}$ .

The diameter of the inscribed sphere, which is twice this, or $6 \sqrt{3}$ , is equal to the length of one edge of the cube.

Either statement alone gives us the length of one side of the cube, which is enough to allow the volume of the pyramid to be calculated.

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Question

Note: Figure NOT drawn to scale.

The above figure shows a rectangular prism with an inscribed tetrahedron, or triangular pyramid, with vertices . What is the volume of the tetrahedron?

Statement 1:

Statement 2:

Answer

We show that the two statements together provide insufficient information by assuming them both to be true.

The volume of the pyramid is one third the product of the height, which is , and the area of the base; this base, being a right triangle, is equal to one half the product of the lengths of its legs, or and . Therefore,

$V = \frac{1}{3} \cdot AE \cdot \frac{1}{2} EF \cdot EH$

or

$V = \frac{1}{6} \cdot AE \cdot EF \cdot EH$

By Statement 1, , and by Statement 2, , so by substitution,

$V = \frac{1}{6} \cdot AE \cdot AE \cdot 40$

$V = \frac{20}{3} \cdot (AE )^{2}$

Without any further information, however, the volume cannot be determined.

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Question

A tetrahedron is a solid with four triangular faces.

Give the volume of a tetrahedron.

Statement 1: The tetrahedron has four equilateral faces.

Statement 2: The surface area of the tetrahedron is $120 \sqrt{3}$ .

Answer

Neither statement is enough to determine the volume of the tetrahedron; Statement 1 alone gives no actual measurements, and Statement 2 gives only the surface area, which can apply to infinitely many tetrahedrons.

Assume both statements to be true. A tetrahedron with four equilateral faces is a regular tetrahedron, whose surface area, relative to the common length of its edges, is defined by the formula

$A = s^{2}\sqrt{3}$ .

By substituting $120 \sqrt{3}$ for , it is possible to calculate . Consequently, the volume of the tetrahedron can be calculated using the volume formula

$V = \frac{s^{3} }{6\sqrt{2}}$ .

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Question

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

Give the volume of a regular tetrahedron.

Statement 1: Each edge has length 8.

Statement 2: Each face has area $16 \sqrt{3}$ .

Answer

The formula for the volume of a regular tetrahedron given the length of each edge is

$V = \frac{s^{3} }{6\sqrt{2}}$ .

Statement 1 gives information explicitly. Statement 2 gives the means to find , since, if $16 \sqrt{3}$ is substituted for in the formula for an equilateral triangle:

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

the value of can be determined.

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure, which shows a tetrahedron, or triangular pyramid. What is the volume of the tetrahedron?

Statement 1: $\bigtriangleup ABD$ is an isosceles triangle with area 64.

Statement 2: $\bigtriangleup ABC$ is an equilateral triangle with perimeter 48.

Answer

Each statement gives enough information about one triangle to determine its area, its angles, and its sidelengths, but no information about the other three triangles is given except for one side.

Assume both statements are known. $\bigtriangleup ABD$ is an isosceles triangle with area 64. Since , we can find this common sidelength using the area formula for a triangle, with these lengths as height and base:

$\frac{1}{2} \cdot AD \cdot BD = 64$

$AD \cdot AD = 128$

$(AD)^{2} = 128$

$AD = \sqrt{128} =8 \sqrt{2}$ .

This is the length of both $\overline{AD}$ and $\overline{BD}$ .

By the 45-45-90 Theorem, $\overline{AB}$ has length $\sqrt{2}$ times this, or $8 \sqrt{2} \cdot \sqrt{2} = 8 \cdot 2 = 16$ .

Since $\bigtriangleup ABC$ is an equilateral triangle, . Since $\bigtriangleup ADC$ is a right triangle, , and $AD =8 \sqrt{2}$ , the triangle is also isosceles, and $CD= 8\sqrt{2}$ ; by a similar argument, $BD= 8\sqrt{2}$ .

The volume of the pyramid can be calculated. Its base, which is congruent to $\bigtriangleup ABD$ , has area 64, and its height is $AD = 8 \sqrt{2}$ ; multiply one third by their product to get the volume.

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Question

In the above diagram, a tetrahedron - a triangular pyramid - with vertices is shown inside a cube. Give the volume of the tetrahedron.

Statement 1: The perimeter of Square is 16.

Statement 2: The area of $\bigtriangleup EGH$ is 8.

Answer

The volume of the pyramid is one third the product of height and the area of its base, which in turn, since here it is a right triangle, is half the product of the lengths and of its legs. Since the prism in the figure is a cube, the three lengths are equal, so we can set each to . The volume of the pyramid is

$V = \frac{1}{3} \cdot s\cdot \frac{1}{2} s\cdot s$

$V = \frac{1}{6} s^{3}$

Therefore, knowing the length of one edge of the cube is sufficient to determine the volume of the pyramid.

Assume Statement 1 alone. Since the perimeter of Square is 16, each side of the square, and each edge of the cube has one fourth this measure, or 4.

Assume Statement 2 alone. $\bigtriangleup EGH$ has congruent legs, each of measure ; since its area is 8, can be found as follows:

$\frac{1}{2} s^{2} = A$

$\frac{1}{2} s^{2} = 8$

$s^{2}= 16$

From either statement alone, the length of each side of the cube, and, subsequently, the volume of the pyramid, can be calculated.

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Question

Note: Figure NOT drawn to scale.

The above figure shows a rectangular prism with an inscribed tetrahedron, or triangular pyramid, with vertices . What is the volume of the tetrahedron?

Statement 1: Isosceles right triangle $\bigtriangleup AEF$ has area 32.

Statement 2:

Answer

The volume of the pyramid is one third the product of the height, which is , and the area of the base; this base, being a right triangle, is equal to one half the product of the lengths of its legs, or and . Therefore,

$V = \frac{1}{3} \cdot AE \cdot \frac{1}{2} EF \cdot EH$

or

$V = \frac{1}{6} \cdot AE \cdot EF \cdot EH$

From Statement 1 alone, we know $\bigtriangleup AEF$ is isosceles and has area 32; therefore, its common leg length can be determined using the area formula:

$\frac{1}{2} s^{2}= 32$

$s^{2}= 64$

Therefore, . However, nothing can be determined about .

Statement 2 alone does not give any of the three desired lengths or any information necessary to find them.

However, Statement 2, along with the information from Statement 1, can be used to find . From Statement 2, , and from Statement 1, ; the Pythagorean Theorem can be used to find . Therefore, all three of , , and can be found, and the volume of the pyramid can be calculated.

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Question

A solid in three-dimensional coordinate space has four vertices, at points , , , and for some positive values of . What is the volume of the solid?

Statement 1:

Statement 2:

Answer

The figure described is the triangular pyramid, or tetrahedron, in the coordinate three-space below.

The base of the pyramid can be seen as a triangle with the three known coordinates , , and , and the area of its base is half the product of the lengths of its legs, which is

$b = \frac{1}{2} \cdot 8 \cdot 12 = 48$ .

The volume of the pyramid is one third the product of the area of its base, which is 48, and its height, which is the perpendicular distance from the unknown point to the base. Since the base is entirely within the -plane, this distance is the -coordinate of the apex, which is . Therefore, the only thing that is needed to determine the volume of the pyramid is ; this information is provided in Statement 2, but not Statement 1.

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Question

Note: Figure NOT drawn to scale.

The above figure shows a rectangular prism with an inscribed tetrahedron, or triangular pyramid, with vertices . What is the volume of the tetrahedron?

Statement 1: Rectangle has area 240.

Statement 2: Square has area 144.

Answer

The volume of the pyramid is one third the product of the height, which is , and the area of the base; this base, being a right triangle, is equal to one half the product of the lengths of its legs, or and . Therefore,

$V = \frac{1}{3} \cdot AE \cdot \frac{1}{2} EF \cdot EH$

or

$V = \frac{1}{6} \cdot AE \cdot EH \cdot EF$

From Statement 1, it can be determined that $AE \cdot EH = 240$ , but without knowing anything about , the volume of the pyramid cannot be determined. Similarly, from Statement 2, it can be determined that $AE \cdot EF = 144$ , but nothing is given about .

Now assume both statements are true. Statement 2 gives that Quadrilateral is a square with area 144, so $AE = EF = \sqrt{144} = 12$ . From Statement 1, we can tell $AE \cdot EH = 240$ . The volume of the pyramid can be calculated as

$V = \frac{1}{6} \cdot AE \cdot EH \cdot EF = \frac{1}{6}\cdot240 \cdot 12= 480$ .

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Question

A solid on the three-dimensional coordinate plane has four vertices, at points , , , and for some positive values of . What is the volume of the solid?

Statement 1:

Statement 2:

Answer

The figure described is the triangular pyramid, or tetrahedron, in the coordinate three-space below.

The base of the pyramid can be seen as a triangle with the three known coordinates , , and , and the area of its base is half the product of the lengths of its legs, which is

$b = \frac{1}{2} \cdot 10 \cdot 14 = 70$ .

The volume of the pyramid is one third the product of the area of its base, which is 70, and its height, which is the perpendicular distance from the unknown point to the base. Since the base is entirely within the -plane, this distance is the -coordinate, which is . Therefore, the only thing that is needed to determine the volume of the pyramid is . However, the two statements together only yield , and therefore do not give sufficient information to solve the problem.

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Question

Note: Figure NOT drawn to scale.

The above figure shows a rectangular prism with an inscribed tetrahedron, or triangular pyramid, with vertices . What is the volume of the tetrahedron?

Statement 1: Rectangle has area 200.

Statement 2: Rectangle has area 120.

Answer

The volume of the pyramid is one third the product of the height, which is , and the area of the base; this base, being a right triangle, is equal to one half the product of the lengths of its legs, or and . Therefore,

$V = \frac{1}{3} \cdot AE \cdot \frac{1}{2} EF \cdot EH$

or

$V = \frac{1}{6} \cdot AE \cdot EF \cdot EH$

We need to know the values of , , and to find the volume of of the pyramid. We show that the two statements give insufficient information by examining two scenarios.

Case 1:

Rectangle has area $AE \cdot EH = 10 \cdot 20 = 200$ .

Rectangle has area $AE \cdot EF = 10 \cdot 12= 120$ .

The volume of the pyramid is

$V = \frac{1}{6} \cdot AE \cdot EF \cdot EH= V = \frac{1}{6} \cdot 10 \cdot 12 \cdot 20 = 400$

Case 2:

Rectangle has area $AE \cdot EH = 8 \cdot 25 = 200$ .

Rectangle has area $AE \cdot EF = 8 \cdot 15= 120$ .

The volume of the pyramid is

$V = \frac{1}{6} \cdot AE \cdot EF \cdot EH= V = \frac{1}{6} \cdot 8 \cdot 25 \cdot 15 =500$

In each case, the conditions of both statements are met, but the volumes of the pyramids differ.

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Question

A solid on the three-dimensional coordinate plane has four vertices, at points , , , and for some positive values of . What is the volume of the solid?

Statement 1:

Statement 2:

Answer

The figure described is the triangular pyramid, or tetrahedron, in the coordinate three-space below.

The base of the pyramid can be seen as a triangle with the three known coordinates , , and , and the area of its base is half the product of the lengths of its legs, which is

$b = \frac{1}{2} \cdot 10 \cdot 14 = 70$ .

The volume of the pyramid is one third the product of the area of its base, which is 70, and its height, which is the perpendicular distance from the unknown point to the base. Since the base is entirely within the -plane, this distance is the -coordinate, which is . Therefore, the only thing that is needed to determine the volume of the pyramid is .

Neither statement alone is enough to gain this information. However, if both statements are assumed true, we can subtract each side of the latter equation from the former as follows:

$\underline{A+B-C= 30}$

The value of is obtained and the volume of the pyramid can be calculated.

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Question

Note: Figure NOT drawn to scale, but you may assume .

In the above figure, a pyramid with a rectangular base is inscribed inside a rectangular prism; its vertices are . What is the volume of the pyramid?

Statement 1: The hypotenuse $\overline{AF}$ of 30-60-90 triangle $\bigtriangleup AEF$ has length 16.

Statement 2: The hypotenuse $\overline{AH}$ of 45-45-90 right triangle $\bigtriangleup AEH$ has length $8\sqrt{2}$ .

Answer

The volume of the pyramid is one third the product of height and the area of its rectangular base, which is $EF \cdot EH$ ; that is,

$V = \frac{1}{3} \cdot AE \cdot EF \cdot EH$

Assume Statement 1 alone. $\bigtriangleup AEF$ is a 30-60-90 triangle with a hypotenuse of length 16. By the 30-60-90 Triangle Theorem, short leg $\overline{AE}$ has length half this, or 8, and long leg $\overline{EF}$ has length $\sqrt{3}$ times that of $\overline{AE}$ , or $8 \sqrt{3}$ . However, the length of $\overline{EH}$ cannot be determined.

Assume Statement 2 alone. $\bigtriangleup AEH$ is a 45-45-90 right triangle with a hypotenuse of length $8\sqrt{2}$ . By the 45-45-90 Theorem, its legs $\overline{AE}$ and $\overline{EH}$ each have length $8\sqrt{2}$ divided by $\sqrt{2}$ , which is $\frac{8\sqrt{2}}{\sqrt{2}}= 8$ ; however, the length of $\overline{EF}$ cannot be determined.

From the two statements together, we can determine that and $EF = 8 \sqrt{3}$ , and calculate the volume:

$V = \frac{1}{3} \cdot 8 \cdot 8 \sqrt{3} \cdot 8 = \frac{512\sqrt{3}}{3}$ .

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Question

Pyramid 1 in three-dimensional coordinate space has as its base the square with vertices at the origin, , , and , and its apex at the point ; Pyramid 2 has as its base the square with vertices at the origin, , , and , and its apex at the point . All six variables represent positive quantities. Which pyramid has the greater volume?

Statement 1: $10 \le F \le 12$ and $10 \le J \le 12$

Statement 2:

Answer

The volume of a pyramid is one third the product of its height and the area its base.

Pyramid 1 is shown below:

The base of the pyramid is on the -plane, so the height of the pyramid is the perpendicular distance from apex to this plane; this is the -coordinate, . The base of the pyramid is a square of sidelength 10, so its area is the square of 10, or 100. This makes the volume of Pyramid 1

$V_{1} = \frac{1}{3} \cdot 100 \cdot F = \frac{100}{3}F = \left ( 33 \frac{1}{3}\right )F$

Similarly, the volume of Pyramid 2 is

$V_{2} = \frac{1}{3} \cdot 12 ^{2}\cdot J= \frac{1}{3} \cdot 144 \cdot J = 36 J$

Therefore, the problem asks us to determine which of $\left ( 33 \frac{1}{3}\right )F$ and is the greater.

We show that the two statements together provide insufficient information by examining two scenarios.

Case 1:

$V_{1} = \left ( 33 \frac{1}{3}\right )F = \left ( 33 \frac{1}{3}\right ) \cdot 12 = 400$

$V_{2} = 36 J = 36 \cdot 10 = 360$

$V_{1} > V_{2}$

Case 2: $F =10 \frac{1}{6}, J = 10$

$V_{1} = \left ( 33 \frac{1}{3}\right )F = \left ( 33 \frac{1}{3}\right ) \cdot 10 \frac{1}{6} = 338 \frac{8}{9}$

$V_{2} = 36 J = 36 \cdot 10 = 360$

$V_{1}< V_{2}$

Each case fits the conditions of the two statements and the main body of the question; in one case, Pyramid 1 has the greater volume and in the other case, Pyramid 2 does.

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Question

Pyramid 1 and Pyramid 2 in three-dimensional coordinate space each have the same base: the square with vertices at the origin, , , and . Pyramid 1 has its fifth vertex at the point ; Pyramid has its fifth vertex at the point . All six variables represent positive quantities. Which pyramid, if either, has the greater volume?

Statement 1:

Statement 2:

Answer

The volume of a pyramid is one third the product of its height and the area its base. The two pyramids have the same base, so the pyramid with the greater height will have the greater volume (and if their heights are equal, their volumes are equal).

Pyramid 1 is shown below:

The base of the pyramid is on the -plane, so the height of the pyramid is the perpendicular distance from apex to this plane; this is the -coordinate, . Pyramid 2 has the same base and apex , so its height is the -coordinate of its apex, .

Therefore, whichever is greater - or - determines which pyramid has the greater volume.

Neither statement alone gives a clue as to which is greater. However, if we assume both, then, by the subtraction property of inequality,

and

together imply that

$D+E+F - ( D+E) > G+H+J - \left ( G+H \right )$

and

.

This means that Pyramid 1 has the greater height and, consequently, the greater volume.

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Question

Pyramid 1 and Pyramid 2 in three-dimensional coordinate space each have the same base: the square with vertices at the origin, , , and . Pyramid 1 has its fifth vertex at the point ; Pyramid 2 has its fifth vertex at the point . All six variables represent positive quantities. Which pyramid, if either, has the greater volume?

Statement 1:

Statement 2:

Answer

The volume of a pyramid is one third the product of its height and the area its base. The two pyramids have the same base, so the pyramid with the greater height will have the greater volume (and if their heights are equal, their volumes are equal).

Pyramid 1 is shown below:

The base of the pyramid is on the -plane, so the height of the pyramid is the perpendicular distance from apex to this plane; this is the -coordinate, . Pyramid 2 has the same base and apex , so its height is the -coordinate of its apex, .

Therefore, whichever is greater - or - determines which pyramid has the greater volume. However, the two statements to not give this information.

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Question

The above figure shows a cube; a pyramid with vertices at is inscribed. What is the volume of the pyramid?

Statement 1: The cube has volume 729.

Statement 2: $\bigtriangleup AGF$ has area $\frac{81 \sqrt{2}}{2}$ .

Answer

Let be the common length of the edges of the cube.

The height of the pyramid is , and the area of its base, Square , is $EF \cdot EH = s^{2}$ . The volume of the pyramid is one third the product of these, or

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

Therefore, if the common length of the edges of the cube can be found, the volume of the pyramid can be calculated.

From Statement 1 alone, this length can be calculated as the cube root of the volume of the cube, 729; this cube root is equal to 9.

From Statement 2 alone, this length can also be calculated. $\bigtriangleup AGF$ has as one leg $\overline{FG}$ with length ; its other leg, which is a diagonal of a square with sidelength has length $s \sqrt{2}$ by the 45-45-90 Theorem. The area is half the product of these legs; since this area is $\frac{81 \sqrt{2}}{2}$ , we can find as follows:

$\frac{1}{2}\cdot s \cdot s\sqrt{2} = \frac{81 \sqrt{2}}{2}$

$\frac{\sqrt{2}}{2} s ^{2} = \frac{81 \sqrt{2}}{2}$

$\frac{2} {\sqrt{2}} \cdot \frac{\sqrt{2}}{2} s ^{2} =\frac{2} {\sqrt{2}} \cdot \frac{81 \sqrt{2}}{2}$

$s^{2} = 81$

From either statement alone, can be calculated to be 9, and the volume of the pyramid can be found to be

$V = \frac{1}{3} s^{3}= \frac{1}{3} \cdot 9^{3} = 81$

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