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.

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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.

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