How to find the surface area of a tetrahedron - PSAT Math

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Question

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

Answer

The area of an equilateral triangle is given by the formula

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

Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is

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

Substitute :

$SA = s^{2}\sqrt{3} = 6^{2}\sqrt{3} = 36\sqrt{3}$ square inches.

Compare your answer with the correct one above

Question

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area

$A = \frac{s^{2} \sqrt{3} }{4} = \frac{7^{2} \sqrt{3} }{4} = \frac{49 \sqrt{3} }{4} \approx \frac{49 \cdot 1.7321}{4} \approx 21.22$

The total surface area is four times this, or about $4 \cdot 21.22 = 84.88$ .

Rounded, this is 84.9.

Compare your answer with the correct one above

Question

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

Answer

The area of an equilateral triangle is given by the formula

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

Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is

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

Substitute :

$SA = s^{2}\sqrt{3} = 6^{2}\sqrt{3} = 36\sqrt{3}$ square inches.

Compare your answer with the correct one above

Question

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area

$A = \frac{s^{2} \sqrt{3} }{4} = \frac{7^{2} \sqrt{3} }{4} = \frac{49 \sqrt{3} }{4} \approx \frac{49 \cdot 1.7321}{4} \approx 21.22$

The total surface area is four times this, or about $4 \cdot 21.22 = 84.88$ .

Rounded, this is 84.9.

Compare your answer with the correct one above

Question

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

Answer

The area of an equilateral triangle is given by the formula

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

Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is

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

Substitute :

$SA = s^{2}\sqrt{3} = 6^{2}\sqrt{3} = 36\sqrt{3}$ square inches.

Compare your answer with the correct one above

Question

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area

$A = \frac{s^{2} \sqrt{3} }{4} = \frac{7^{2} \sqrt{3} }{4} = \frac{49 \sqrt{3} }{4} \approx \frac{49 \cdot 1.7321}{4} \approx 21.22$

The total surface area is four times this, or about $4 \cdot 21.22 = 84.88$ .

Rounded, this is 84.9.

Compare your answer with the correct one above

Question

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

Answer

The area of an equilateral triangle is given by the formula

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

Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is

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

Substitute :

$SA = s^{2}\sqrt{3} = 6^{2}\sqrt{3} = 36\sqrt{3}$ square inches.

Compare your answer with the correct one above

Question

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area

$A = \frac{s^{2} \sqrt{3} }{4} = \frac{7^{2} \sqrt{3} }{4} = \frac{49 \sqrt{3} }{4} \approx \frac{49 \cdot 1.7321}{4} \approx 21.22$

The total surface area is four times this, or about $4 \cdot 21.22 = 84.88$ .

Rounded, this is 84.9.

Compare your answer with the correct one above

Question

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

Answer

The area of an equilateral triangle is given by the formula

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

Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is

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

Substitute :

$SA = s^{2}\sqrt{3} = 6^{2}\sqrt{3} = 36\sqrt{3}$ square inches.

Compare your answer with the correct one above

Question

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area

$A = \frac{s^{2} \sqrt{3} }{4} = \frac{7^{2} \sqrt{3} }{4} = \frac{49 \sqrt{3} }{4} \approx \frac{49 \cdot 1.7321}{4} \approx 21.22$

The total surface area is four times this, or about $4 \cdot 21.22 = 84.88$ .

Rounded, this is 84.9.

Compare your answer with the correct one above

Question

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

Answer

The area of an equilateral triangle is given by the formula

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

Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is

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

Substitute :

$SA = s^{2}\sqrt{3} = 6^{2}\sqrt{3} = 36\sqrt{3}$ square inches.

Compare your answer with the correct one above

Question

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area

$A = \frac{s^{2} \sqrt{3} }{4} = \frac{7^{2} \sqrt{3} }{4} = \frac{49 \sqrt{3} }{4} \approx \frac{49 \cdot 1.7321}{4} \approx 21.22$

The total surface area is four times this, or about $4 \cdot 21.22 = 84.88$ .

Rounded, this is 84.9.

Compare your answer with the correct one above

Question

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

Answer

The area of an equilateral triangle is given by the formula

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

Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is

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

Substitute :

$SA = s^{2}\sqrt{3} = 6^{2}\sqrt{3} = 36\sqrt{3}$ square inches.

Compare your answer with the correct one above

Question

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area

$A = \frac{s^{2} \sqrt{3} }{4} = \frac{7^{2} \sqrt{3} }{4} = \frac{49 \sqrt{3} }{4} \approx \frac{49 \cdot 1.7321}{4} \approx 21.22$

The total surface area is four times this, or about $4 \cdot 21.22 = 84.88$ .

Rounded, this is 84.9.

Compare your answer with the correct one above

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How to find the surface area of a tetrahedron - PSAT Math

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle. A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

The area of an equilateral triangle is given by the formula Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is Substitute : square inches.

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area The total surface area is four times this, or about . Rounded, this is 84.9.

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle. A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

The area of an equilateral triangle is given by the formula Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is Substitute : square inches.

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area The total surface area is four times this, or about . Rounded, this is 84.9.

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle. A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

The area of an equilateral triangle is given by the formula Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is Substitute : square inches.

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area The total surface area is four times this, or about . Rounded, this is 84.9.

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle. A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

The area of an equilateral triangle is given by the formula Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is Substitute : square inches.

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area The total surface area is four times this, or about . Rounded, this is 84.9.

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle. A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

The area of an equilateral triangle is given by the formula Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is Substitute : square inches.

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area The total surface area is four times this, or about . Rounded, this is 84.9.

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle. A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

The area of an equilateral triangle is given by the formula Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is Substitute : square inches.

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area The total surface area is four times this, or about . Rounded, this is 84.9.

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle. A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

The area of an equilateral triangle is given by the formula Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is Substitute : square inches.

Give the surface area of the above tetrahedron, or four-faced solid, to the nearest tenth.

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area The total surface area is four times this, or about . Rounded, this is 84.9.

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length six inches. Give the total surface area of the tetrahedron.