Geometry (B): READ: Surface Area of Prisms Reading

READ: Surface Area of Prisms Reading

Prisms

Learning Objectives

Use nets to represent prisms.
Find the surface area of a prism.
Find the volume of a prism.

Introduction

A prism is a three-dimensional figure with a pair of parallel and congruent ends, or bases. The sides of a prism are parallelograms. Prisms are identified by their bases.

Surface Area of a Prism Using Nets

The prisms above are right prisms. In a right prism, the lateral sides are perpendicular to the bases of prism. Compare a right prism to an oblique prism, in which sides and bases are not perpendicular.

Two postulates that apply to area are the Area Congruence Postulate and the Area Addition Postulate.

Area Congruence Postulate:

If two polygons (or plane figures) are congruent, then their areas are congruent.

Area Addition Postulate:

The surface area of a three-dimensional figure is the sum of the areas of all of its non-overlapping parts.

You can use a net and the Area Addition Postulate to find the surface area of a right prism.

From the net, you can see that that the surface area of the entire prism equals the sum of the figures that make up the net:

$\text{Total surface area} = \text{area} \ A + \text{area} \ B + \text{area} \ C + \text{area} \ D + \text{area} \ E + \text{area} \ F$

Using the formula for the area of a rectangle, you can see that the area of rectangle $A$ is:

$A & = l \cdot w\\ A & = 10 \cdot 5 = 50 \ \text{square units}$

Similarly, the areas of the other rectangles are inserted back into the equation above.

$\text{Total surface area} & = \text{area} \ A + \text{area} \ B + \text{area} \ C + \text{area} \ D + \text{area} \ E + \text{area} \ F\\ \text{Total surface area} & = (10 \cdot 5) + (10 \cdot 3) + (10 \cdot 5) + (10 \cdot 3) + (5 \cdot 3) + (5 \cdot 3)\\ \text{Total surface area} & = 50 + 30 + 50 + 30 + 15 + 15\\ \text{Total surface area} & = 190 \ \text{square units}$

Example 9

Use a net to find the surface area of the prism.

The area of the net is equal to the surface area of the figure. To find the area of the triangle, we use the formula:

$A = 1/2 \;\mathrm{hb}$ where $h$ is the height of the triangle and $b$ is its base.

Note that triangles $A$ and $E$ are congruent so we can multiply the area of triangle $A$ by $2$ .

$\text{area} &= \text{area }A +\text{area }B+\text{area }C+\text{area }D+\text{area }E \\ &=2(\text{area }A) + \text{area }B+\text{area }C+\text{area }D \\ &=2\left [\frac{1}{2}(9 \cdot 12)\right ]+(6 \cdot 9) +(6 \cdot 12) + (6 \cdot 15) \\ &= 108 + 54 + 72 + 90 \\ &= 324$

Thus, the surface area is $324 \;\mathrm{square\ units}$ .

Surface Area of a Prism Using Perimeter

This hexagonal prism has two regular hexagons for bases and six sides. Since all sides of the hexagon are congruent, all of the rectangles that make up the lateral sides of the three-dimensional figure are also congruent. You can break down the figure like this.

The surface area of the rectangular sides of the figure is called the lateral area of the figure. To find the lateral area, you could add up all of the areas of the rectangles.

$\text{lateral area} &= 6 \ \text{(area of one rectangle)} \\ &=6 \ (s \cdot h) \\ &= 6sh$

Notice that $6s$ is the perimeter of the base. So another way to find the lateral area of the figure is to multiply the perimeter of the base by $h$ , the height of the figure.

$\text{lateral area } &= 6sh \\ &=(6s )\cdot h \\ &= (\text{perimeter})h \\ &= Ph$

Substituting $P$ , the perimeter, for $6s$ , we get the formula for any lateral area of a right prism:

$\text{lateral area of a prism} = Ph$

Now we can use the formula to calculate the total surface area of the prism. Using $P$ for the perimeter and $B$ for the area of a base:

$\text{Total surface area} & = \text{lateral area} + \text{area of 2 bases}\\ & = \text{(perimeter of base} \cdot \text{height)} + 2 \ \text{(area of base)}\\ & = \text{Ph} + 2 \ B$

To find the surface area of the figure above, first find the area of the bases. The regular hexagon is made of six congruent small triangles. The altitude of each triangle is the apothem of the polygon. Note: be careful here—we are talking about the altitude of the triangles, not the height of the prism. We find the length of the altitude of the triangle using the Pythagorean Theorem, $a = \sqrt{4^{2}-2^{2}} \approx 3.46$

So the area of each small triangle is:

$A \ \text{(triangle)} &= \frac{1}{2} ab\\ &= \frac{1}{2} (3.46)(4)\\ &= 6.92$

The area of the entire hexagon is therefore:

$A \ \text{(base)} &= 6 \ \text{(area of triangle)}\\ &= 6 \cdot 6.92\\ &= 41.52$

You can also use the formula for the area of a regular polygon to find the area of each base:

$A \ \text{(polygon)} &= \frac{1}{2} aP\\ &= \frac{1}{2} (3.46)(24)\\ &= 41.52$

Now just substitute values to find the surface area of the entire figure above.

$\text{(total surface area)} &= Ph + 2B\\ &= [(6 \cdot 4) \cdot 10] + 2 (41.52)\\ &= (24 \cdot 10) + 83.04\\ &= 240 + 83.04\\ &= 323.04 \ \text{square units}$

You can use the formula $A = Ph + 2B$ to find the surface area of any right prism.

Example 10

Use the formula to find the total surface area of the trapezoidal prism.

The dimensions of the trapezoidal base are shown. Set up the formula. We’ll call the height of the entire prism $H$ to avoid confusion with $h$ , the height of each trapezoidal base.

$\text{Total surface area} = PH + 2B$

Now find the area of each trapezoidal base. You can do this by using the formula for the area of a trapezoid. (Note that the height of the trapezoid, $2.46$ is small $h$ .)

$A &= \frac{1}{2} h(b_1 + b_2)\\ &= \frac{1}{2} (2.64)[10 + 4]\\ &= 18.48 \ \text{square units}$

Now find the perimeter of the base.

$P &= 10 + 4 + 4 + 4\\ &= 22$

Now find the total surface area of the solid.

$\text{(total surface area)} &= Ph + 2B\\ &= (22)(21) + 2(18.48)\\ &= 462 + 36.96\\ &= 498.96 \ \text{square units}$

The following review questions are for your own benefit. The answers are listed below for you to check your work and understanding.

Review Questions

For each of the following find the surface area using

a. the method of nets and

b. the perimeter.

The base of a prism is a right triangle whose legs are $3$ and $4$ and show height is $20$ . What is the total area of the prism?
A right hexagonal prism is $24\;\mathrm{inches}$ tall and has bases that are regular hexagons measuring $8\;\mathrm{inches}$ on a side. What is the total surface area?

For problems 6 and 7:

A barn is shaped like a pentagonal prism with dimensions shown in feet:

How many square feet (excluding the roof) are there on the surface of the barn to be painted?
If a gallon of paint covers $250\;\mathrm{square\ feet}$ , how many gallons of paint are needed to paint the barn?

Review Answers

$40.5 \;\mathrm{in}^2$
$838 \;\mathrm{cm}^2$
$252\;\mathrm{square\ units}$
$1484.6\;\mathrm{square\ units}$
$2450\;\mathrm{square\ feet}$
$10\;\mathrm{gallons}$ of paint

Last modified: Wednesday, 30 June 2010, 9:15 AM

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