Calculating the surface area of a tetrahedron - GMAT Quantitative

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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?

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

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.

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

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?

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.

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?

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?

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.

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.

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.

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

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?

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.

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 $\small (0,0,0), (60,0,0), (0,60,0), (60,0,0)$ .

Give the surface area of the tetrahedron.

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

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

Compare your answer with the correct one above

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?

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 $\small 400 \sqrt{3}$ .

Compare your answer with the correct one above

Question

Refer to the above diagram, which shows a tetrahedron.

, and . Give the surface area of the tetrahedron.

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?

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?

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.

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.

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.

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

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?

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.

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 $\small (0,0,0), (60,0,0), (0,60,0), (60,0,0)$ .

Give the surface area of the tetrahedron.

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

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

Compare your answer with the correct one above

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