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Math 8 · Mathematics
Geometric solids · Geometric solids

Surface Area of Polyhedra

Before you start, make sure you know: Pythagorean Theorem (Math 7)

Emma is building a cardboard model of the Luxor, the pyramid-shaped hotel in Las Vegas, and she also needs to wrap a gift box. How much cardboard does the model need? How much wrapping paper does the box need? Both questions ask for the surface area of a polyhedron: the sum of the areas of all its faces. The trick is to unfold the solid and see it flat.

Nets and surface area

If you cut along some edges of a polyhedron and flatten it onto a table, you get its net. The net shows every face at once, so the surface area of the solid is just the area of its net.

Net of a cube made of six equal squares, with the surface area formula and an example for an edge of 4 inches. net of a cube: 6 squares Surface area SA = 6 · s² Edge 4 in.: 6 · 4² = 96 in.²

Areas are measured in square units (in.², ft², cm²), because you are adding up flat regions.

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A young designer is creating a smart speaker prototype in the shape of a cube. If the edge of the cube measures $9\text{ cm}$, calculate the total surface area of…

A young designer is creating a smart speaker prototype in the shape of a cube. If the edge of the cube measures $9\text{ cm}$, calculate the total surface area of plastic material needed to cover all its faces.

9 in 9 in 9 in
Solve
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From the net of the figure, identify which polyhedron it is and calculate its total area knowing that each square has a side of $5\text{ cm}$.

From the net of the figure, identify which polyhedron it is and calculate its total area knowing that each square has a side of $5\text{ cm}$.

5 in 5 in
Solve

Surface area of a prism

When you unroll the lateral faces of a right prism, they form one big rectangle. Its width is the perimeter of the base and its height is the height of the prism.

Net of a right prism: the lateral faces form one long rectangle whose length is the perimeter of the base, with two bases attached. perimeter of the base h Lateral area L = P · h Surface area SA = L + 2B

Sometimes you are given the volume instead of the height. Recall $V = B \cdot h$, so $h = V \div B$. Find $h$ first, then use $L = P \cdot h$.

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A shoe box has the shape of a rectangular prism with the following dimensions: $30\text{ cm}$ long, $18\text{ cm}$ wide and $12\text{ cm}$ high. Calculate the total area…

A shoe box has the shape of a rectangular prism with the following dimensions: $30\text{ cm}$ long, $18\text{ cm}$ wide and $12\text{ cm}$ high. Calculate the total area of cardboard used to manufacture the box (excluding flaps).

30 in 12 in 18 in
Solve
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A musical instrument case has the shape of a right triangular prism. The base is an equilateral triangle of side $8\text{ cm}$ and the height of the prism is…

A musical instrument case has the shape of a right triangular prism. The base is an equilateral triangle of side $8\text{ cm}$ and the height of the prism is $40\text{ cm}$. If the area of the triangular base is approximately $27,7\text{ cm}^2$, calculate the total area of the case.

Triangular base A ≈ 27.7 in² h = 40 in 8 in
Solve
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Calculate the lateral area of a regular pentagonal prism knowing that its volume is $450\text{ cm}^3$, its base area is $30\text{ cm}^2$ and the side of the base…

Calculate the lateral area of a regular pentagonal prism knowing that its volume is $450\text{ cm}^3$, its base area is $30\text{ cm}^2$ and the side of the base measures $4\text{ cm}$.

(Hint: First use the volume formula to find the height of the prism).

4 cm Given V = 450 cm³ B = 30 cm² l = 4 cm h = ? Lat. Area = ?
Solve

Regular polygon bases

Many prisms have a regular polygon as a base: a pentagon, a hexagon, an octagon. The apothem $a$ of a regular polygon is the distance from its center to the midpoint of a side. To find the area of the base, split the polygon into $n$ identical triangles. Each triangle has base $s$ (the side) and height $a$.

$$B = n \cdot \tfrac{1}{2} s a = \tfrac{1}{2} P a$$

When the apothem is not given, find it with the Pythagorean Theorem. In a regular hexagon the distance from the center to a vertex equals the side $s$, so the right triangle below has legs $a$ and $s/2$ and hypotenuse $s$.

Regular hexagon with a right triangle formed by the apothem, half a side, and the segment from the center to a vertex. a s/2 s Right triangle: a² + (s/2)² = s² so a = (√3 / 2) · s

If the problem gives the perimeter and the apothem directly, skip the Pythagorean step and plug them into $B = \tfrac{1}{2} P a$.

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For a technology project, you have to build a toolbox in the shape of a regular hexagonal prism. The edge of the base measures $10\text{ in}$ and the height of the prism…

For a technology project, you have to build a toolbox in the shape of a regular hexagonal prism. The edge of the base measures $10\text{ in}$ and the height of the prism is $25\text{ in}$. If you want to cover all the outer faces (including the two bases) with a protective film, how many square inches of material will you need?

(Note: Remember that to calculate the area of the base you will first need to find the apothem of the hexagon using the Pythagorean theorem).

a 10 5 25 in 10 in
Solve
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A metal bin without a lid is to be manufactured in the shape of a regular octagonal prism. The base has a perimeter of $64\text{ cm}$ and an apothem of $9,66\text{ cm}$.…

A metal bin without a lid is to be manufactured in the shape of a regular octagonal prism. The base has a perimeter of $64\text{ cm}$ and an apothem of $9,66\text{ cm}$. The height of the bin is $40\text{ cm}$. Calculate the total surface area of metal needed for its manufacture.

Without lid ap=9.66 h=40 Base perimeter = 64 in
Solve

Rhombus and trapezoid bases

For these bases the lateral faces are rectangles, but you must first find the side lengths of the base. Both cases come down to a right triangle.

  • Rhombus. The diagonals cross at right angles and cut each other in half. Each side is the hypotenuse of a right triangle with legs equal to half of each diagonal. The area is $B = \tfrac{1}{2} d_1 d_2$.

  • Isosceles trapezoid. Drop a perpendicular from the top corner. The horizontal leg is $\dfrac{\text{long base} - \text{short base}}{2}$ and the vertical leg is the height. The slanted side is the hypotenuse. The area is $B = \tfrac{1}{2}(b_1 + b_2)\,h$.

A rhombus with diagonals 16 and 12 and side 10, and an isosceles trapezoid with bases 14 and 6, height 3, and slanted side 5. 8 6 10 rhombus 14 6 4 3 5 isosceles trapezoid

Some problems mix units, for example inches for the base, meters for the height, and an answer in square feet. The rule is always the same: convert every length to the unit of the answer first, then compute. Here are the facts you need:

Conversion Use
$1 \text{ ft} = 12 \text{ in.}$ inches → feet: divide by 12
$1 \text{ m} \approx 3.28 \text{ ft}$ meters → feet: multiply by 3.28
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Calculate the lateral area of a right prism whose height is $12\text{ cm}$ and whose base is a rhombus with diagonals $16\text{ cm}$ and $12\text{ cm}$. (Hint: First you…

Calculate the lateral area of a right prism whose height is $12\text{ cm}$ and whose base is a rhombus with diagonals $16\text{ cm}$ and $12\text{ cm}$.

(Hint: First you must find the side of the rhombus using the Pythagorean theorem in one of the interior right triangles of the base).

8 6 l h = 12 in 6 in 8 in l l² = 6² + 8²
Solve
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A right prism of height $20\text{ cm}$ has an isosceles trapezoid as its base. The parallel sides of the trapezoid measure $14\text{ cm}$ and $6\text{ cm}$, and its…

A right prism of height $20\text{ cm}$ has an isosceles trapezoid as its base. The parallel sides of the trapezoid measure $14\text{ cm}$ and $6\text{ cm}$, and its height is $3\text{ cm}$. Calculate the total area of the prism. Hint: You will first need to calculate the length of the slanted sides of the base trapezoid using the Pythagorean theorem.

Solve
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An industrial waste container has been manufactured in the shape of a right prism whose base is a rhombus. The diagonals of the base rhombus measure $160\text{ in}$ and…

An industrial waste container has been manufactured in the shape of a right prism whose base is a rhombus. The diagonals of the base rhombus measure $160\text{ in}$ and $120\text{ in}$. If the height of the container is $2\text{ meters}$, calculate the total outer surface area of the container in square feet ($m^2$).

Solve

The space diagonal of a box

The space diagonal of a rectangular prism joins two opposite corners and passes through the inside of the box. You find it by using the Pythagorean Theorem twice: first on the floor, then in a vertical right triangle.

A rectangular box with length, width, and height labeled and its space diagonal drawn from one bottom corner to the opposite top corner. length l w h d Space diagonal d² = l² + w² + h² d = √(l² + w² + h²)

Why? The floor diagonal $f$ satisfies $f^2 = l^2 + w^2$. The space diagonal $d$ is the hypotenuse of a right triangle with legs $f$ and $h$, so $d^2 = f^2 + h^2 = l^2 + w^2 + h^2$.

When the diagonal is given and a dimension is missing, plug in what you know and solve for the unknown. If the problem says one dimension is twice another, write both in terms of the same letter before you substitute.

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An architecture studio is designing a loft in the shape of a rectangular prism. It is known that the length of the room is twice its width, and that the height is…

An architecture studio is designing a loft in the shape of a rectangular prism. It is known that the length of the room is twice its width, and that the height is exactly $3\text{ m}$. If the space diagonal (the maximum distance between two opposite corners) measures exactly $7\text{ m}$, calculate the total area of the walls and the ceiling (excluding the floor) that must be covered with insulating paint.

D = 7 ft 2x x 3 ft
Solve

Surface area of a pyramid

The lateral faces of a regular pyramid are identical triangles. Each has base $s$ (a side of the base) and height $\ell$, the slant height of the pyramid. The slant height is the distance from the apex to the midpoint of a base edge, measured along a face. Some problems call it the apothem of the pyramid.

$$L = \frac{P \cdot \ell}{2} \qquad\qquad SA = L + B$$ Net of a square pyramid: a square base with four triangles around it, and the slant height marked on one triangle. ℓ 4 triangles: ½ · s · ℓ each + base: s²

The slant height $\ell$ is not the height $h$ of the pyramid. The height goes straight down to the center of the base. If you are given $h$, build a right triangle with legs $h$ and $s/2$ (half the base edge):

Cross section of a square pyramid showing the right triangle with height h, half of the base edge, and slant height. h s/2 ℓ Slant height ℓ² = h² + (s/2)²

A skylight shaped like a pyramid has no base (it is open to the room), so for glass you only need $L$. A closed pyramid needs $SA = L + B$.

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Calculate the lateral area of a regular square-based pyramid if we know that the side of the base measures $10\text{ cm}$ and the apothem of the pyramid (the height of…

Calculate the lateral area of a regular square-based pyramid if we know that the side of the base measures $10\text{ cm}$ and the apothem of the pyramid (the height of its lateral faces) measures $15\text{ cm}$.

Ap = 15 in l = 10 in
Solve
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An architecture company has designed a glass skylight in the shape of a regular square-based pyramid. The edge of the base measures $4\text{ m}$ and the height of the…

An architecture company has designed a glass skylight in the shape of a regular square-based pyramid. The edge of the base measures $4\text{ m}$ and the height of the pyramid is $3\text{ m}$. Calculate the total surface area of glass needed to cover the four lateral faces (the base is left open to let light into the building).

4 ft h = 3 ft 2 ft a = ?
Solve

Surface area of a frustum

A frustum of a pyramid is what remains when you slice off the top of a pyramid with a cut parallel to the base. Its two bases are similar polygons, and its lateral faces are trapezoids. For a regular square frustum, each trapezoid has area $\tfrac{1}{2}(s_B + s_b)\,\ell$, where $\ell$ is the slant height of the frustum. Adding the four faces gives

$$L = \frac{(P_B + P_b)\cdot \ell}{2} \qquad\qquad SA = L + B_{\text{big}} + B_{\text{small}}$$

If you are given the vertical height $h$ instead of $\ell$, draw the side view. The horizontal leg of the right triangle is $\dfrac{s_B - s_b}{2}$.

Side view of a frustum: a trapezoid with a right triangle showing the height, half the difference of the bases, and the slant height. (sB − sb)/2 h ℓ small base big base Slant height ℓ² = h² + ((sB − sb)/2)²
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Calculate the total area of a frustum of a regular square-based pyramid whose bases are squares with sides $12\text{ cm}$ and $6\text{ cm}$ respectively, and whose…

Calculate the total area of a frustum of a regular square-based pyramid whose bases are squares with sides $12\text{ cm}$ and $6\text{ cm}$ respectively, and whose lateral face height (slant height of the frustum) is $8\text{ cm}$. What is the minimum amount of wrapping paper needed to wrap an object of this shape?

12 in 6 in ap = 8 in
Solve
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A luxury perfume brand has designed a bottle in the shape of a frustum of a regular square-based pyramid. The edge of the larger base measures $8\text{ cm}$ and the edge…

A luxury perfume brand has designed a bottle in the shape of a frustum of a regular square-based pyramid. The edge of the larger base measures $8\text{ cm}$ and the edge of the smaller base measures $4\text{ cm}$. If the height of the bottle is $10\text{ cm}$, calculate the total surface area of glass needed to manufacture the bottle (including both bases). To do this, you will first need to calculate the apothem of the frustum using the Pythagorean theorem in 3D space.

Solve

Regular polyhedra

A regular polyhedron has faces that are all the same regular polygon, with the same number of faces meeting at every vertex. There are only five of them.

Solid Faces Shape of a face Vertices Edges
Tetrahedron 4 triangle 4 6
Cube 6 square 8 12
Octahedron 8 triangle 6 12
Dodecahedron 12 pentagon 20 30
Icosahedron 20 triangle 12 30

Since all faces are identical:

$$SA = (\text{number of faces}) \times (\text{area of one face})$$

For an equilateral triangle with side $s$, the height is $\sqrt{s^2 - (s/2)^2} = \tfrac{\sqrt{3}}{2}s$. So the area is $\tfrac{1}{2} \cdot s \cdot \tfrac{\sqrt{3}}{2}s$:

$$A_{\text{triangle}} = \frac{\sqrt{3}}{4}\, s^2$$

If a problem gives a price per square inch (for example, silver plating), multiply the surface area by that price to get the cost.

When an answer should be "in terms of $\sqrt{3}$", leave the radical in the answer instead of using a decimal, as in $288\sqrt{3}$.

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A glazier must manufacture a glass for a lamp in the shape of a regular tetrahedron (all its faces are equal equilateral triangles). If each face has an area of…

A glazier must manufacture a glass for a lamp in the shape of a regular tetrahedron (all its faces are equal equilateral triangles). If each face has an area of $15,6\text{ cm}^2$, how much glass surface area will the complete lamp have?

1 face = 15.6 in² Total Area = ?
Solve
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A trophy has the shape of a regular octahedron with an edge of $6\text{ in}$. The trophy is going to be silver-plated, and the cost of the plating is $0,05\text{ \$}…

A trophy has the shape of a regular octahedron with an edge of $6\text{ in}$. The trophy is going to be silver-plated, and the cost of the plating is $0,05\text{ \$}$ per square inch. What will be the total cost of plating the $8$ faces of the trophy?

(Given: The area of an equilateral triangle of side $a$ is $A = \frac{\sqrt{3}}{4} a^2$).

a = 6 in A = (√3/4)a² a = 6 in
Solve
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An ant walks over the surface of a regular octahedron with edge $12\text{ cm}$. If the octahedron can be considered as the union of two regular square-based pyramids…

An ant walks over the surface of a regular octahedron with edge $12\text{ cm}$. If the octahedron can be considered as the union of two regular square-based pyramids joined at their base, calculate the total area the ant can cover. To do this, remember that all the faces of a regular octahedron are equilateral triangles. Express the result in terms of $\sqrt{3}$.

Solve

Euler's Formula

Count the parts of any convex polyhedron: $F$ faces, $V$ vertices (corners), and $E$ edges. A surprising pattern appears.

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Verify if Euler's theorem ($C + V = A + 2$) holds for a pentagonal prism. Clearly state the number of faces ($C$), vertices ($V$) and edges ($A$) you have counted.

Verify if Euler's theorem ($C + V = A + 2$) holds for a pentagonal prism. Clearly state the number of faces ($C$), vertices ($V$) and edges ($A$) you have counted.

Faces (C) = 7 Vertices (V) = 10 Edges (A) = 15 C + V = A + 2 7 + 10 = 15 + 2
Solve
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A new polyhedron for a board game has been designed complying with Euler's theorem. We know that the die has $14$ vertices and that all its faces are equilateral…

A new polyhedron for a board game has been designed complying with Euler's theorem. We know that the die has $14$ vertices and that all its faces are equilateral triangles. Could you deduce how many faces this polyhedron has and what the total number of edges is? Prove your answer using Euler's mathematical relationship.

Vertex (V) Face (C) Edge (A) V = 14 C = ? A = ? Euler's formula C + V = A + 2 3·C = 2·A (Triangular faces)
Solve
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"Euler's theorem" states that in convex polyhedra $C + V = A + 2$. Imagine a polyhedron that has the same number of faces as vertices. If the number of edges is $12$,…

"Euler's theorem" states that in convex polyhedra $C + V = A + 2$. Imagine a polyhedron that has the same number of faces as vertices. If the number of edges is $12$, deduce how many faces it has and suggest the name of a regular polyhedron that meets this condition. Once identified, calculate its total area if its edge measures $\sqrt{10}\text{ cm}$.

Euler's theorem C + V = A + 2 a = √10 in Vertices (V = 7) Faces (C = 7) Edges (A = 12) Hexagonal pyramid
Solve

Combined and cut solids

Sometimes a solid is built by gluing two solids together, or by cutting a piece off or out. For the surface area, ask one question: which faces are on the outside?

  • When you glue two solids along a face, that face is hidden. Do not count it for either solid.

  • When you cut a piece out, the faces that disappear are subtracted, and the new faces that appear on the cut are added.

Side view of a pyramid glued on top of a prism, with the shared dashed face marked as hidden. Glued face is hidden: it is not on the outside of the prism, and not of the pyramid. count: prism sides + prism bottom + pyramid sides
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We are designing a decorative wooden piece consisting of a right square-based prism with base $8\text{ cm}$ and height $10\text{ cm}$ from which a pyramid with the same…

We are designing a decorative wooden piece consisting of a right square-based prism with base $8\text{ cm}$ and height $10\text{ cm}$ from which a pyramid with the same base and the same height has been removed. Calculate the total surface area of this piece (including the inner "hollow" left by the pyramid).

10 in 10 in 8 in Pyramidal hollow
Solve
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A metal part is to be built by joining a regular hexagonal prism and a regular hexagonal pyramid that fits perfectly on its top base. The side of the base hexagon…

A metal part is to be built by joining a regular hexagonal prism and a regular hexagonal pyramid that fits perfectly on its top base. The side of the base hexagon measures $6\text{ cm}$, the height of the prism is $15\text{ cm}$ and the height of the pyramid is $8\text{ cm}$. Calculate the outer surface area of the complete part (bear in mind that the base where both solids join does not count as an outer surface).

8 in aₚ a 15 in
Solve
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Consider a cube of edge $10\text{ cm}$. If we make a planar cut passing through the midpoints of three edges that meet at the same vertex, we obtain a small triangular…

Consider a cube of edge $10\text{ cm}$. If we make a planar cut passing through the midpoints of three edges that meet at the same vertex, we obtain a small triangular pyramid (called a truncated tetrahedron). Calculate the total area of the solid that remains after removing that small pyramid at that corner.

5 in 5 in 5 in 5 in 10 in New face (equilateral triangle)
Solve

Back to the start: For Emma's gift box (a closed prism) she needs the surface area, lateral area plus two bases. For the pyramid model with an open bottom she needs only the lateral area, built from triangles whose height is the slant height.

Summary

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