DSQ: Calculating the surface area of a tetrahedron - GMAT Quantitative

Card 0 of 16

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

Tap the card to reveal the answer