How to find the surface area of a polyhedron

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PSAT Math › How to find the surface area of a polyhedron

Questions 1 - 2

1

A regular octahedron has eight congruent faces, each of which is an equilateral triangle.

A given octahedron has edges of length five inches. Give the total surface area of the octahedron.

$50\sqrt{3} \textrm{ in}^{2}$

$25\sqrt{3} \textrm{ in}^{2}$

$50\sqrt{2} \textrm{ in}^{2}$

$25\sqrt{2} \textrm{ in}^{2}$

$50 \textrm{ in}^{2}$

Explanation

The area of an equilateral triangle is given by the formula

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

Since there are eight equilateral triangles that comprise the surface of the octahedron, the total surface area is

$SA = 8 A = 8 \cdot \frac{s^{2}\sqrt{3}}{4} = 2 s^{2}\sqrt{3}$

Substitute :

$SA =2 \cdot 5^{2}\sqrt{3} = 50\sqrt{3}$ square inches.

2

A regular icosahedron has twenty congruent faces, each of which is an equilateral triangle.

The total surface area of a given regular icosahedron is 400 square centimeters. To the nearest tenth of a centimeter, what is the length of each edge?

$6.8\textrm{ cm}$

$13.6\textrm{ cm}$

$11.9\textrm{ cm}$

$23.8\textrm{ cm}$

$17.8\textrm{ cm}$

Explanation

The total surface area of the icosahedron is 400 square centimeters; since the icosahedron comprises twenty congruent faces, each has area $400 \div 20 = 20$ square centimeters.

The area of an equilateral triangle is given by the formula

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

Set and solve for :

$\frac{s^{2}\sqrt{3}}{4} = 20$

$\frac{s^{2}\sqrt{3}}{4} \cdot \frac{4}{\sqrt{3}}= 20 \cdot \frac{4}{\sqrt{3}}$

$s^{2} = \frac{80}{ \sqrt{3}} \approx \frac{80}{1.732} \approx 46.19$

$s \approx \sqrt{46.19} \approx 6.8$ centimeters.

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