Chapter 10: Perimeter and Area

Triangles and Parallelograms

Learning Objectives

Understand basic concepts of the meaning of area.
Use formulas to find the area of specific types of polygons.

Introduction

Measurement is not a new topic. You have been measuring things nearly all your life. Sometimes you use standard units (pound, centimeter), sometimes nonstandard units (your pace or arm span). Space is measured according to its dimension.

One-dimensional space: measure the length of a segment on a line.
Two-dimensional space: measure the area that a figure takes up on a plane (flat surface).
Three-dimensional space: measure the volume that a solid object takes up in “space.”

In this lesson, we will focus on basic ideas about area in two-dimensional space. Once these basic ideas are established we’ll look at the area formulas for some of the most familiar two-dimensional figures.

Basic Ideas of Area

Measuring area is just like measuring anything; before we can do it, we need to agree on standard units. People need to say, “These are the basic units of area.” This is a matter of history. Let’s re-create some of the thinking that went into decisions about standard units of area.

Example 1

What is the area of the rectangle below?

What should we use for a basic unit of area?

As one possibility, suppose we decided to use the space inside this circle as the unit of area.

To find the area, you need to count how many of these circles fit into the rectangle, including parts of circles.

So far you can see that the rectangle’s space is made up of $8$ whole circles. Determining the fractional parts of circles that would cover the remaining white space inside the rectangle would be no easy job! And this is just for a very simple rectangle. The challenge is even more difficult for more complex shapes.

Instead of filling space with circles, people long ago realized that it is much simpler to use a square shape for a unit of area. Squares fit together nicely and fill space with no gaps. The square below measures $1 \;\mathrm{foot}$ on each side, and it is called $1 \;\mathrm{square \ foot}$ .

Now it’s an easy job to find the area of our rectangle.

The area is $8 \;\mathrm{square \ feet}$ , because $8$ is the number of units of area (square feet) that will exactly fill, or cover, the rectangle.

The principle we used in Example 1 is more general.

The area of a two-dimensional figure is the number of square units that will fill, or cover, the figure.

Two Area Postulates

Congruent Areas

If two figures are congruent, they have the same area.

This is obvious because congruent figures have the same amount of space inside them. However, two figures with the same area are not necessarily congruent.

Area of Whole is Sum of Parts

If a figure is composed of two or more parts that do not overlap each other, then the area of the figure is the sum of the areas of the parts.

This is the familiar idea that a whole is the sum of its parts. In practical problems you may find it helpful to break a figure down into parts.

Example 2

Find the area of the figure below.

Luckily, you don’t have to learn a special formula for an irregular pentagon, which this figure is. Instead, you can break the figure down into a trapezoid and a triangle, and use the area formulas for those figures.

Basic Area Formulas

Look back at Example 1 and the way it was filled with unit area squares.

Notice that the dimensions are:

base (or length) $4 \;\mathrm{feet}$

height (or width) $2 \;\mathrm{feet}$

But notice, too, that the base is the number of feet in one row of unit squares, and the height is the number of rows. A counting principle tells us that the total number of square feet is the number in one row multiplied by the number of rows.

$\text{Area} = 8 = 4 \times 2 = \text{base} \times \text{height}$

Area of a Rectangle

If a rectangle has base $b$ units and height $h$ units, then the area, $A$ , is $bh$ square units.

$A = bh$

Example 3

What is the area of the figure shown below?

Break the figure down into two rectangles.

$\text{Area} = 22 \times 45 + 8 \times 20 = 990 + 160 = 1150 \;\text{cm}^2$

Now we can build on the rectangle formula to find areas of other shapes.

Parallelogram

Example 4

How could we find the area of this parallelogram?

Make it into a rectangle

The rectangle is made of the same parts as the parallelogram, so their areas are the same. The area of the rectangle is $bh$ , so the area of the parallelogram is also $bh$ .

Warning: Notice that the height $h$ of the parallelogram is the perpendicular distance between two parallel sides of the parallelogram, not a side of the parallelogram (unless the parallelogram is also a rectangle, of course).

Area of a Parallelogram

If a parallelogram has base $b$ units and height $h$ units, then the area, $A$ , is $bh$ square units.

$A = bh$

Triangle

Example 5

How could we find the area of this triangle?

Make it into a parallelogram. This can be done by making a copy of the original triangle and putting the copy together with the original.

The area of the parallelogram is $bh$ , so the area of the triangle is $\frac{bh}{2}$ or $\frac{1}{2}bh.$

Warning: Notice that the height $h$ (also often called the altitude) of the triangle is the perpendicular distance between a vertex and the opposite side of the triangle.

Area of a Triangle

If a triangle has base $b$ units and altitude $h$ units, then the area, $A$ , is $\frac{bh}{2}$ or $\frac{1}{2}$ $bh$ square units.

$A = \frac{bh}{2}$ or $A = \frac{1}{2} bh$

Lesson Summary

Once we understood the meaning of measures of space in two dimensions—in other words, area—we saw the advantage of using square units. With square units established, the formula for the area of a rectangle is simply a matter of common sense. From that point forward, the formula for the area of each new figure builds on the previous figure. For a parallelogram, convert it to a rectangle. For a triangle, double it to make a parallelogram.

Points to Consider

As we study other figures, we will frequently return to the basics of this lesson—the benefit of square units, and the fundamental formula for the area of a rectangle.

It might be interesting to note that the word geometry is derived from ancient Greek roots that mean Earth (geo-) measure (-metry). In ancient times geometry was very similar to today’s surveying of land. You can see that land surveying became easily possible once knowledge of how to find the area of plane figures was developed.

Review Questions

Complete the chart. Base and height are given in units; area is in square units.

	Base	Height	'Area
1a.	$5$	$8$	?
1b.	$10$	?	$40$
1c.	$1$	$1$	?
1d.	$7$	?	$49$
1e.	$225$	$\frac{1}{3}$	?
1f.	$100$	?	$1$

The carpet for a $12-\mathrm{foot}$ by $20-\mathrm{foot}$ room cost $\$360$ . The same kind of carpet cost $\$225$ for a room with a square floor. What are the dimensions of the room?
Explain how an altitude of a triangle can be outside the triangle.
Line $k$ and line $m$ are parallel.

Explain how you know that $\triangle ABX , \triangle ABY$ , and $\triangle ABZ$ all have the same area.
Lin bought a tract of land for a new apartment complex. The drawing below shows the measurements of the sides of the tract. Approximately how many acres of land did Lin buy? $(1\;\mathrm{acre} \approx 40,000\;\mathrm{square\ feet}.)$
A hexagon is drawn on a coordinate grid. The vertices of the hexagon are $A(1, 4), B(3, 7), C(8, 7), D(6, 4), E(6, -4),$ and $F(1, -8).$ What is the area of $ABDCEF?$

Review Answers

1a. $40$
1b. $4$

1c. $1$

1d. $7$

1e. $75$

1f. $0.01$
$15 \;\mathrm{feet}$ by $15 \;\mathrm{feet}$
This happens in a triangle with an obtuse angle. Each altitude to a side of the obtuse angle is outside the triangle.
All of the triangles have the same base and altitude, so in each triangle $\frac{bh}{2}$ is the same as in each of the other triangles.
$160,000 + 420,000 + 280,000 = 860,000\;\mathrm{sq \ ft} \approx 21.5\;\mathrm{acres}$
$65$

Trapezoids, Rhombi, and Kites

Learning Objectives

Understand the relationships between the areas of two categories of quadrilaterals: basic quadrilaterals (rectangles and parallelograms), and special quadrilaterals (trapezoids, rhombi, and kites).
Derive area formulas for trapezoids, rhombi, and kites.
Apply the area formulas for these special quadrilaterals.

Introduction

We’ll use the area formulas for basic shapes to work up to the formulas for special quadrilaterals. It’s an easy job to convert a trapezoid to a parallelogram. It’s also easy to take apart a rhombus or kite and rebuild it as a rectangle. Once we do this, we can derive new formulas from the old ones.

We’ll also need to review basic facts about the trapezoid, rhombus, and kite.

Area of a Trapezoid

Recall that a trapezoid is a quadrilateral with one pair of parallel sides. The lengths of the parallel sides are the bases. The perpendicular distance between the parallel sides is the height, or altitude, of the trapezoid.

To find the area of the trapezoid, turn the problem into one about a parallelogram. Why? Because you already know how to compute the area of a parallelogram.

Make a copy of the trapezoid.
Rotate the copy $180^\circ$ .
Put the two trapezoids together to form a parallelogram.

Two things to notice:

The parallelogram has a base that is equal to $b_1 + b_2$ .
The altitude of the parallelogram is the same as the altitude of the trapezoid.

Now to find the area of the trapezoid:

The area of the parallelogram is $\mathrm{base} \times\;\mathrm{altitude} = (b_1 + b_2) \times h$ .
The parallelogram is made up of two congruent trapezoids, so the area of each trapezoid is one-half the area of the parallelogram.
The area of the trapezoid is one-half of $(b_1 + b_2) \times h$ .

Area of Trapezoid with Bases $b_1$ and $b_2$ and Altitude $h$

Trapezoid with bases $b_1$ and $b_2$ and altitude $h$

$A = \frac{1} {2} (b_1 + b_2) h$ or $A = \frac{(b_1 + b_2)h} {2}$

Notice that the formula for the area of a trapezoid could also be written as the "Average of the bases time the height." This may be a convenient shortcut for memorizing this formula.

Example 1

What is the area of the trapezoid below?

The bases of the trapezoid are $4$ and $6$ . The altitude is $3$ .

$A=\frac{1}{2}({b_1+b_2})h=\frac{1}{2}({4+6})\times 3=15$

Area of a Rhombus or Kite

First let’s start with a review of some of the properties of rhombi and kites.

	Kite	Rhombus
Congruent sides	$2$ Pairs	All $4$
Opposite angles congruent	$1$ Pair yes. $1$ Pair maybe	Both pairs yes
Perpendicular diagonals	Yes	Yes
Diagonals bisected	$1$ Yes. $1$ maybe	Both yes

Now you’re ready to develop area formulas. We’ll follow the command: “Frame it in a rectangle.” Here’s how you can frame a rhombus in a rectangle.

Notice that:

The base and height of the rectangle are the same as the lengths of the two diagonals of the rhombus.
The rectangle is divided into $8$ congruent triangles; $4$ of the triangles fill the rhombus, so the area of the rhombus is one-half the area of the rectangle.

Area of a Rhombus with Diagonals $d_1$ and $d_2$

$A=\frac{1}{2}{d_1d_2}=\frac{d_1d_2}{2}$

We can go right ahead with the kite. We’ll follow the same command again: “Frame it in a rectangle.” Here’s how you can frame a kite in a rectangle.

Notice that:

The base and height of the rectangle are the same as the lengths of the two diagonals of the kite.
The rectangle is divided into $8$ triangles; $4$ of the triangles fill the kite. For every triangle inside the kite, there is a congruent triangle outside the kite so the area of the kite is one-half the area of the rectangle.

Area of a Kite with Diagonals $d_1$ and $d_2$

$A = \frac{1}{2}{d_1d_2} = \frac{d_1d_2}{2}$

Lesson Summary

We see the principle of “no need to reinvent the wheel” in developing the area formulas in this section. If we wanted to find the area of a trapezoid, we saw how the formula for a parallelogram gave us what we needed. In the same way, the formula for a rectangle was easy to modify to give us a formula for rhombi and kites. One of the striking results is that the same formula works for both rhombi and kites.

Points to Consider

You’ll use area concepts and formulas later in this course, as well as in real life.

Surface area of solid figures: the amount of outside surface.
Geometric probability: chances of throwing a dart and landing in a given part of a figure.
Carpet for floors, paint for walls, fertilizer for a lawn, and more: areas needed.

Tech Note - Geometry Software

You saw earlier that the area of a rhombus or kite depends on the lengths of the diagonals.

$A=\frac{1}{2}{d_1d_2}=\frac{d_1d_2}{2}$

This means that all rhombi and kites with the same diagonal lengths have the same area.

Try using geometry software to experiment as follows.

Construct two perpendicular segments.
Adjust the segments so that one or both of the segments are bisected.
Draw a quadrilateral that the segments are the diagonals of. In other words, draw a quadrilateral for which the endpoints of the segments are the vertices.
Repeat with the same perpendicular, bisected segments, but making a different rhombus or kite. Repeat for several different rhombi and kites.
Regardless of the specific shape of the rhombus or kite, the areas are all the same.

The same activity can be done on a geoboard. Place two perpendicular rubber bands so that one or both are bisected. Then place another rubber band to form a quadrilateral with its vertices at the endpoints of the two segments. A number of different rhombi and kites can be made with the same fixed diagonals, and therefore the same area.

Review Questions

Quadrilateral $ABCD$ has vertices $A(-2, 0), B(0, 2), C(4, 2),$ and $D(0, -2)$ in a coordinate plane.

Show that $ABCD$ is a trapezoid.
What is the area of $ABCD$ ?
Prove that the area of a trapezoid is equal to the area of a rectangle with height the same as the height of the trapezoid and base equal to the length of the median of the trapezoid.
Show that the trapezoid formula can be used to find the area of a parallelogram.
Sasha drew this plan for a wood inlay he is making.

$10$ is the length of the slanted side. $16$ is the length of the horizontal line segment. Each shaded section is a rhombus.

The shaded sections are rhombi. Based on the drawing, what is the total area of the shaded sections?
Plot on a coordinate plane.
- The points are the vertices of a rhombus.
- The area of the rhombus is $24 \;\mathrm{square\ units}$ .
Tyra designed the logo for a new company. She used three congruent kites.

What is the area of the entire logo?
In the figure below:
- $ABCD$ is a square
- $AP = PB = BQ$
- $DC = 20\;\mathrm{feet}$
What is the area of $PBQC$ ?

In the figure below:
- $ABCD$ is a square
- $AP = 20\;\mathrm{feet}$
- $PB = BQ = 10\;\mathrm{feet}$
What is the area of $PBQC$ ?
The area of $PBQD$ is what fractional part of the area of $ABCD$ ?

Review Answers

Slope of $\overline{A B} = 1$ , slope of $\overline{D C} = 1$

$\overline{AB}\parallel\overline{DC}$ are parallel.

$AB&=\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\ DC&=\sqrt{4^2+4^2}=\sqrt{32}=4\sqrt{2}\ AD&=\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}$

slope of $\overline{A D} = -1$

$\overline{A B}$ and $\overline{D C}$ are the bases, $\overline{A D}$ is an altitude.

$A = \frac{(b_1 + b_2)h} {2} = \frac{(2\sqrt{2} + 4\sqrt{2})2\sqrt{2}} {2} = \frac{6\sqrt{2}(2\sqrt{2})} {2} = 12$

.
For a parallelogram, $b_1$ = $b_2 = b$ (the “bases” are two of the parallel sides), so by the trapezoid formula the area is
$\frac{(b_1 + b_2)h} {2} = \frac{(b + b)h} {2} = \frac{2bh} {2} = bh.$
Length of long diagonal of one rhombus is $16$ . Length of other diagonal is $12$ (each rhombus is made of $4 6-8-10$ right triangles).

Total area is $2\left [ \frac{d_1d_2}{2} \right ]=d_1d_2=16\times12=192.$

Many rhombi work, as long as the product of the lengths of the diagonals is $48$ .
$90\;\mathrm{cm}^2$
$200\;\mathrm{square\ feet}$
$300\;\mathrm{square\ feet}$
$\frac{1}{3}$

Areas of Similar Polygons

Learning Objectives

Understand the relationship between the scale factor of similar polygons and their areas.
Apply scale factors to solve problems about areas of similar polygons.
Use scale models or scale drawings.

Introduction

We’ll begin with a quick review of some important features of similar polygons. You remember that we studied similar figures rather extensively in Chapter 7. There you learned about scale factors and perimeters of similar polygons. In this section we’ll take similar figures one step farther. We’ll see that the areas of similar figures have a very specific relationship to the scale factor—but it’s just a bit tricky! We wrap up the section with some thoughts on why living things are the “right” size, and what geometry has to do with that!

Review - Scale Factors and Perimeter

Example 1

The diagram below shows two rhombi.

a. Are the rhombi similar? How do you know?

Yes.

The sides are parallel, so the corresponding angles are congruent.
Using the Pythagorean Theorem, we can see that each side of the smaller rhombus has a length of $10$ , and each side of the larger rhombus has a length of $15$ .
So the lengths of the sides are proportional.
Polygons with congruent corresponding angles and proportional sides are similar.

b. What is the scale factor relating the rhombi?

The scale factor relating the smaller rhombus to the larger one is $\frac{15}{10} = \frac{3}{2}=1.5.$

c. What is the perimeter of each rhombus?

Answer

$\mathrm{Perimeter\ of\ smaller\ rhombus} = 4 \times 10 = 40$
$\mathrm{Perimeter\ of\ larger\ rhombus} = 4 \times 15 = 60$

d. What is the ratio of the perimeters?

$\frac{60}{40} = \frac {3}{2}=1.5$

e. What is the area of each rhombus?

$\text{Area of smaller rhombus} & = \frac{d_1d_2}{2}=\frac {{12}\times{16}}{2}= 96 \ \text{Area of larger rhombus} & = \frac{d_1d_2}{2}=\frac {{18}\times{24}}{2}= 216$

What do you notice in this example? The perimeters have the same ratio as the scale factor.

But what about the areas? The ratio of the areas is certainly not the same as the scale factor. If it were, the area of the larger rhombus would be $96 \times 1.5 = 144$ , but the area of the larger rhombus is actually $216.$

What IS the ratio of the areas?

The ratio of the areas is $\frac{216}{96} = \frac {9}{6}=2.25.$ Notice that $\frac{9}{4}=\left ( \frac{3}{2} \right)^2$ or in decimal, $2.25 = (1.5)^2$ .

So at least in this case we see that the ratio of the areas is the square of the scale factor.

Scale Factors and Areas

What happened in Example 1 is no accident. In fact, this is the basic relationship for the areas of similar polygons.

Areas of Similar Polygons

If the scale factor relating the sides of two similar polygons is $k$ , then the area of the larger polygon is $k^2$ times the area of the smaller polygon. In symbols let the area of the smaller polygon be $A_1$ and the area of the larger polygon be $A_2$ . Then:

$A_2 = k^2 A_1$

Think about the area of a polygon. Imagine that you look at a square with an area of exactly $1 \;\mathrm{square\ unit}.$ Of course, the sides of the square are $1 \;\mathrm{unit}$ of length long. Now think about another polygon that is similar to the first one with a scale factor of $k$ . Every $1-$ by $-1$ square in the first polygon has a matching $k-$ by $-k$ square in the second polygon, and the area of each of these $k-$ by $-k$ squares is $k^2$ . Extending this reasoning, every $1 \;\mathrm{square\ unit}$ of area in the first polygon has a corresponding $k^2$ units of area in the second polygon. So the total area of the second polygon is $k^2$ times the area of the first polygon.

Warning: In solving problems it’s easy to forget that you do not always use just the scale factor. Use the scale factor in problems about lengths. But use the square of the scale factor in problems about area!

Example 2

Wu and Tomi are painting murals on rectangular walls. The length and width of Tomi’s wall are $3$ times the length and width of Wu’s wall.

a. The total length of the border of Tomi’s wall is $120 \;\mathrm{feet}$ . What is the total length of the border of Wu’s wall?

This is a question about lengths, so you use the scale factor itself. All the sides of Tomi’s wall are $3$ times the length of the corresponding side of Wu’s wall, so the perimeter of Tomi’s wall is also $3$ times the perimeter of Wu’s wall.

The total length of the border (perimeter) of Wu’s wall is $\frac{120}{3} = 40 \;\mathrm{feet}$ .

b. Wu can cover his wall with $6$ quarts of paint. How many quarts of paint will Tomi need to cover her wall?

This question is about area, since the area determines the amount of paint needed to cover the walls. The ratio of the amounts of paint is the same as the ratio of the areas (which is the square of the scale factor). Let $x$ be the amount of paint that Tomi needs.

$\frac{x}{6} & = k^2=3^2=9 \ x & = 6 \times 9 = 54$

Tomi would need $54$ quarts of paint.

Summary of Length and Area Relationships for Similar Polygons

If two similar polygons are related by a scale factor of $k$ , then:

Length: The lengths of any corresponding parts have the same ratio, $k$ . Note that this applies to sides, *Area: The ratio of the areas is $k^2$ . Note that this applies to areas, and any aspect of an object that

Note: You might be able to make a pretty good guess about the volumes of similar solid $(3-D)$ figures. You’ll see more about that in Chapter 11.

Scale Drawings and Scale Models

One important application of similar figures is the use of scale drawings and scale models. These are two-dimensional (scale drawings) or three-dimensional (scale models) representations of real objects. The drawing or model is similar to the actual object.

Scale drawings and models are widely used in design, construction, manufacturing, and many other fields. Sometimes a scale is shown, such as “ $1\;\mathrm{inch}= 5\;\mathrm{miles}$ ” on a map. Other times the scale may be calculated, if necessary, from information about the object being modeled.

Example 3

Jake has a map for a bike tour. The scale is $1\;\mathrm{inch} = 5\;\mathrm{miles}$ . He estimated that two scenic places on the tour were about $3\frac{1}{2}\;\mathrm{inches}$ apart on the map. How far apart are these places in reality?

Each inch on the map represents a distance of $5\;\mathrm{miles}$ . The places are about $3\frac{1}{2}\times 5 =17.5\;\mathrm{miles}$ apart.

Example 4

Cristy’s design team built a model of a spacecraft to be built. Their model has a scale of $1:24$ . The actual spacecraft will be $180\;\mathrm{feet}$ long. How long should the model be?

Let $x$ be the length of the model.

$\frac{1}{24}& =\frac {x}{180} \ 24x & = 180 \ x & = 7.5$

The model should be $7.5 \;\mathrm{feet}$ long.

Example 5

Tasha is making models of several buildings for her senior project. The models are all made with the same scale. She has started the chart below.

a. What is the scale of the models?

$1250 \div 20 = 62.5$

The scale is $1\;\mathrm{inch} = 62.5\;\mathrm{feet}$ .

b. Complete the chart below.

Building	Actual height (feet)	Model height (inches)
Sears Tower (Chicago)	?	$23.2$
Empire State Building (New York City)	$1250$	$20$
Columbia Center (Seattle)	$930$	?

Sears Tower: $23.2 \times 62.5 = 1450$ . It is $1450\;\mathrm{feet}$ high.

Columbia Center: $\mathrm{Let}\ x = \mathrm{the\ model\ height}$ .

$\frac{1250}{20}& =\frac {930}{x} \ 1250x & = 20 \times 930 \ x & =\frac{{20}\times{930}}{1250}\thickapprox 14.9$

The model should be about $14.9\;\mathrm{inches}$ high.

Why There Are No 12-Foot-Tall Giants

Why are there no $12-\mathrm{foot}-$ tall giants? One explanation for this is a matter of similar figures.

Let’s suppose that there is a $12-\;\mathrm{foot}-$ tall human. Compare this giant (?) to a $6-\;\mathrm{foot}-$ tall person. Now let’s apply some facts about similar figures.

The scale factor relating these two hypothetical people is $\frac{12}{6}=2$ . Here are some consequences of this scale factor.

All linear dimensions of the giant would be $2$ times the corresponding dimensions of the real person. This includes height, bone length, etc.
All area measures of the giant would be $2^2 = 4$ times the corresponding area measures of the real person. This includes respiration (breathing) and metabolism (converting nutrients to usable materials and energy) rates, because these processes take place along surfaces in the lungs, intestines, etc. This also includes the strength of bones, which depends on the cross-section area of the bone.
All volume measures of the giant would be $2^3 = 8$ times the corresponding volume measures of the real person. (You’ll learn why in Chapter 11.) The volume of an organism generally determines its weight and mass.

What kinds of problems do we see for our giant? Here are two severe ones.

The giant would have bones that are $4$ times as strong, but those bones have to carry a body weight that is $8$ times as much. The bones would not be up to the task. In fact it appears that the giant’s own weight would be able to break its bones.
The giant would have $8$ times the weight, number of cells, etc. of the real person, but only $4$ times as much ability to supply the oxygen, nutrition, and energy needed.

Conclusion: There are no $12-\;\mathrm{foot}-$ giants, and some of the reasons are nothing more, or less, than the geometry of similar figures.

For further reading: On Being the Right Size, by J. B. S. Haldane, also available at http://irl.cs.ucla.edu/papers/right-size.html.

Lesson Summary

In his lesson we focused on one main point: The areas of similar polygons have a ratio that is the square of the scale factor. We also used ideas about similar figures to analyze scale drawings and scale models, which are actually similar representations of actual objects.

Points to Consider

You have now learned quite a bit about the lengths of sides and areas of polygons. Next we’ll build on knowledge about polygons to come to a conclusion about the “perimeter” of the “ultimate polygon,” which is the circle.

Suppose we constructed regular polygons that are all inscribed in the same circle.

Think about polygons that have more and more sides.
How would the perimeter of the polygons change as the number of sides increases?

The answers to these questions will lead us to an understanding of the formula for the circumference (perimeter) of a circle.

Review Questions

The figure below is made from small congruent equilateral triangles.

$4$ congruent small triangles fit together to make a bigger, similar triangle.

What is the scale factor of the large and small triangles?
If the area of the large triangle is $20\;\mathrm{square\ units}$ , what is the area of a small triangle?
The smallest squares in the diagram below are congruent.
What is the scale factor of the shaded square and the largest square?
If the area of the shaded square is $50\;\mathrm{square\ units}$ , what is the area of he largest square?
Frank drew two equilateral triangles. Each side of one triangle is $2.5$ times as long as a side of the other triangle. The perimeter of the smaller triangle is $40\;\mathrm{cm}$ . What is the perimeter of the larger triangle?
In the diagram below, . $\overline{M N}: \overline{P Q}$ .
What is the scale factor of the small triangle and the large triangle?
If the perimeter of the large triangle is $42$ , what is the perimeter of the small triangle?
If the area of the small triangle is $A$ , write an expression for the area of the large triangle.
If the area of the small triangle is $K$ , write an expression for the area of the trapezoid.
The area of one square on a game board is exactly twice the area of another square. Each side of the larger square is $50\;\mathrm{mm}$ long. How long is each side of the smaller square?
The distance from Charleston to Morgantown is $160\;\mathrm{miles}$ . The distance from Fairmont to Elkins is $75\;\mathrm{miles}$ . Charleston and Morgantown are $5\;\mathrm{inches}$ apart on a map. How far apart are Fairmont and Elkins on the same map?

Marlee is making models of historic locomotives (train engines). She uses the same scale for all of her models.

The $S1$ locomotive was $140 \;\mathrm{feet}$ long. The model is $8.75 \;\mathrm{inches}$ long.
The $520$ Class locomotive was $87\;\mathrm{feet}$ long.

What is the scale of Marlee’s models?
How long is the model of the $520$ Class locomotive?

Review Answers

$2$
$5$
$\frac{4}{9}$ or $4: 9$
$112.5$
$100\;\mathrm{cm}$
$\frac{2}{3}$
$28$
$\frac{9}{4}A$ or $\frac{9A}{4}$
$\frac{5}{4}K$ or $\frac{5K}{4}$
$35.4\;\mathrm{mm}$
$2.3\;\mathrm{inches}$
$1\;\mathrm{inch}= 16\;\mathrm{feet}$ or equivalent
$5.4\;\mathrm{inches}$

Circumference and Arc Length

Learning Objectives

Understand the basic idea of a limit.
Calculate the circumference of a circle.
Calculate the length of an arc of a circle.

Introduction

In this lesson, we extend our knowledge of perimeter to the perimeter—or circumference—of a circle. We’ll use the idea of a limit to derive a well-known formula for the circumference. We’ll also use common sense to calculate the length of part of a circle, known as an arc.

The Parts of a Circle

A circle is the set of all points in a plane that are a given distance from another point called the center. Flat round things, like a bicycle tire, a plate, or a coin, remind us of a circle.

The diagram reviews the names for the “parts” of a circle.

The center
The circle: the points that are a given distance from the center (which does not include the center or interior)
The interior: all the points (including the center) that are inside the circle
circumference: the distance around a circle (exactly the same as perimeter)
radius: any segment from the center to a point on the circle (sometimes “radius” is used to mean the length of the segment and it is usually written as $r$ )
diameter: any segment from a point on the circle, through the center, to another point on the circle (sometimes “diameter” is used to mean the length of the segment and it is usually written as $d$ )

If you like formulas, you can already write one for a circle:

$d = 2r$ or $(r = \frac {d}{2})$

Circumference Formula

The formula for the circumference of a circle is a classic. It has been known, in rough form, for thousands of years. Let’s look at one way to derive this formula.

Start with a circle with a diameter of $1\;\mathrm{unit}$ . Inscribe a regular polygon in the circle. We’ll inscribe regular polygons with more and more sides and see what happens. For each inscribed regular polygon, the perimeter will be given (how to figure that is in a review question).

What do you notice?

The more sides there are, the closer the polygon is to the circle itself.
The perimeter of the inscribed polygon increases as the number of sides increases.
The more sides there are, the closer the perimeter of the polygon is to the circumference of the circle.

Now imagine that we continued inscribing polygons with more and more sides. It would become nearly impossible to tell the polygon from the circle. The table below shows the results if we did this.

Regular Polygons Inscribed in a Circle with Diameter $1$

Number of sides of polygon	Perimeter of polygon
$3$	$2.598$
$4$	$2.828$
$5$	$2.939$
$6$	$3.000$
$8$	$3.062$
$10$	$3.090$
$20$	$3.129$
$50$	$3.140$
$100$	$3.141$
$500$	$3.141$

As the number of sides of the inscribed regular polygon increases, the perimeter seems to approach a “limit.” This limit, which is the circumference of the circle, is approximately $3.141$ . This is the famous and well-known number $\pi$ . $\pi$ is an endlessly non-repeating decimal number. We often use $\pi \approx 3.14$ as a value for $\pi$ in calculations, but this is only an approximation.

Conclusion: The circumference of a circle with diameter $1$ is $\pi$ .

For Further Reading

Mathematicians have calculated the value of $\pi$ to thousands, and even millions, of decimal places. You might enjoy finding some of these megadecimal numbers. Of course, all are approximately equal to $3.14$ .

The article at the following URL shows more than a million digits of the decimal for $\pi$ . http://wiki.answers.com/Q/What_is_the_exact_value_for_Pi_at_this_moment

Tech Note - Geometry Software

You can use geometry software to continue making more regular polygons inscribed in a circle with diameter $1$ and finding their perimeters.

Can we extend this idea to other circles? First, recall that all circles are similar to each other. (This is also true for all equilateral triangles, all squares, all regular pentagons, etc.)

Suppose a circle has a diameter of $d \;\mathrm{units}$ .

The scale factor of this circle and the one in the diagram and table above, with diameter $1$ , is $d: 1,$ , $\frac{d}{1}$ , or just $d$ .
You know how a scale factor affects linear measures, which include perimeter and circumference. If the scale factor is $d$ , then the perimeter is $d$ times as much.

This means that if the circumference of a circle with diameter $1$ is $\pi$ , then the circumference of a circle with diameter $d$ is $\pi d$ .

Circumference Formula

Let $d$ be the diameter of a circle, and $C$ the circumference.

$C = \pi d$

Example 1

A circle is inscribed in a square. Each side of the square is $10\;\mathrm{cm}$ long. What is the circumference of the circle?

Use $C = \pi d$ . The length of a side of the square is also the diameter of the circle. $C = \pi d = 10\pi \approx 31.4 \;\text{ cm}$

Note that sometimes an approximation is given using $\pi \approx 3.14$ . In this example the circumference is $31.4\;\mathrm{cm}$ using that approximation. An exact is given in terms of $\pi$ (leaving the symbol for $\pi$ in the answer rather than multiplying it out. In this example the exact circumference is $10 \pi\;\mathrm{cm}$ .

Arc Length

Arcs are measured in two different ways.

Degree measure: The degree measure of an arc is the fractional part of a $360^\circ$ complete circle that the arc is.
Linear measure: This is the length, in units such as centimeters and feet, if you traveled from one end of the arc to the other end.

Example 2

Find the length of $\widehat{PQ}$ .

$m\widehat{PQ}$ = $60^\circ$ . The radius of the circle is $9\;\mathrm{inches}$ .

Remember, $60^\circ$ is the measure of the central angle associated with $m\widehat{PQ}$ .

$m\widehat{PQ}$ is $\frac{60}{360}$ of a circle. The circumference of the circle is

$\pi d = 2\pi r = 2\pi(9) = 18\pi\;\mathrm{inches}$ . The arc length of $PQ:$ is $\frac{60}{360}\times 18\pi=\frac {1}{6}\times 18\pi=3 \pi \approx 9.42\;\mathrm{inches}$ .

In this lesson we study the second type of arc measure—the measure of an arc’s length. Arc length is directly related to the degree measure of an arc.

Suppose a circle has:

circumference $C$
diameter $d$
radius $r$

Also, suppose an arc of the circle has degree measure $m$ .

Note that $\frac{m}{360}$ is the fractional part of the circle that the arc represents.

Arc length

$\text{Arc Length}=\frac{m}{360}\times c=\frac {m}{360}\times \pi d=\frac {m}{360}\times 2\pi r$

Lesson Summary

This lesson can be summarized with a list of the formulas developed.

Radius and diameter: $d = 2 r$
Circumference of a circle: $C = \pi d$
$\mathrm{Arc\ length} = \frac{m}{360}\times c=\frac {m}{360}\times \pi d=\frac {m}{360}\times 2\pi r$

Points to Consider

After perimeter and circumference, the next logical measure to study is area. In this lesson, we learned about the perimeter of a circle (circumference) and the arc length of a sector. In the next lesson we’ll learn about the areas of circles and sectors.

Review Questions

Prove: The circumference of a circle with radius $r$ is $2\pi r.$
The Olympics symbol is five congruent circles arranged as shown below. Assume the top three circles are tangent to each other.

Brad is tracing the entire symbol for a poster. How far will his pen point travel?

A truck has tires that measure from the center of the wheel to the outer edge of the tire.
1. How far forward does the truck travel every time a tire turns exactly once?
2. How many times will the tire turn when the truck travels $1\;\mathrm{mile}$ ? $(1\;\mathrm{mile} = 5280\;\mathrm{feet})$ .
The following wire sculpture was made from two perpendicular segments that intersect each other at the center of a circle.
1. If the radius of the circle is $25\;\mathrm{cm}$ , how much wire was used to outline the shaded sections?
The circumference of a circle is $300\;\mathrm{feet}$ . What is the radius of the circle?
A gear with a radius of $3\;\mathrm{inches}$ inches turns at a rate of $2000$ RPM (revolutions per minute). How far does a point on the edge of the pulley travel in one second?
A center pivot irrigation system has a boom that is $400\;\mathrm{m}$ long. The boom is anchored at the center pivot. It revolves around the center pivot point once every three days. How far does the tip of the boom travel in one day?
The radius of Earth at the Equator is about . Belem (in Brazil) and the Galapagos Islands (in the Pacific Ocean) are on (or very near) the Equator. The approximate longitudes are Belem, , and Galapagos Islands, .
1. What is the degree measure of the major arc on the Equator from Belem to the Galapagos Islands?
2. What is the distance from Belem to the Galapagos Islands on the Equator the “long way around?”
A regular polygon inscribed in a circle with diameter $1$ has $n$ sides. Write a formula that expresses the perimeter, $p$ , of the polygon in terms of $n$ . (Hint: Use trigonometry.)
The pulley shown below revolves at a rate of RPM.
1. How far does point $A$ travel in one hour?

Review Answers

$C = \pi d , d = 2r, C = \pi(2r) = 2\pi r$
$40 \pi \approx 125.6\;\mathrm{inches}$
1. $28 \pi \approx 87.92\;\mathrm{inches}$
2. Approximately $721$ times
$100 + 25\pi \approx 178.5\;\mathrm{cm}$
Approximately $47.8\;\mathrm{feet}$
Approximately $628\;\mathrm{inches}$
Approximately $837\;\mathrm{m}$
1. $320^\circ$
2. Approximately $22,329\;\mathrm{miles}$
$p=n \sin \left ( \frac{180}{n} \right )$ or equivalent
$480,000 \pi \approx 1,507,200\;\mathrm{cm}$

Circles and Sectors

Learning Objectives

Calculate the area of a circle.
Calculate the area of a sector.
Expand understanding of the limit concept.

Introduction

In this lesson we complete our area toolbox with formulas for the areas of circles and sectors. We’ll start with areas of regular polygons, and work our way to the limit, which is the area of a circle. This may sound familiar; it’s exactly the same approach we used to develop the formula for the circumference of a circle.

Area of a Circle

The big idea:

Find the areas of regular polygons with radius $1$ .
Let the polygons have more and more sides.
See if a limit shows up in the data.
Use similarity to generalize the results.

The details:

Begin with polygons having $3, 4,$ and $5$ sides, inscribed in a circle with a radius of $1$ .

Now imagine that we continued inscribing polygons with more and more sides. It would become nearly impossible to tell the polygon from the circle. The table below shows the results if we did this.

Regular Polygons Inscribed in a Circle with Radius $1$

Number of sides of polygon	Area of polygon
$3$	$1.2990$
$4$	$2.0000$
$5$	$2.3776$
$6$	$2.5981$
$8$	$2.8284$
$10$	$2.9389$
$20$	$3.0902$
$50$	$3.1333$
$100$	$3.1395$
$500$	$3.1415$
$1000$	$3.1416$
$2000$	$3.1416$

As the number of sides of the inscribed regular polygon increases, the area seems to approach a “limit.” This limit is approximately $3.1416$ , which is $\pi$ .

Conclusion: The area of a circle with radius $1$ is $\pi$ .

Now we extend this idea to other circles. You know that all circles are similar to each other.

Suppose a circle has a radius of $r \;\mathrm{units}$ .

The scale factor of this circle and the one in the diagram and table above, with radius $1$ , is $r: 1, \frac{r}{1}$ , or just $r$ .
You know how a scale factor affects area measures. If the scale factor is $r$ , then the area is $r^2$ times as much.

This means that if the area of a circle with radius $1$ is $\pi$ , then the area of a circle with radius $r$ is $\pi r^2$ .

Area of a Circle Formula

Let $r$ be the radius of a circle, and $A$ the area.

$A=\pi r^2$

You probably noticed that the reasoning about area here is very similar to the reasoning in an earlier lesson when we explored the perimeter of polygons and the circumference of circles.

Example 1

A circle is inscribed in a square. Each side of the square is $10\;\mathrm{cm}$ long. What is the area of the circle?

Use $A=\pi r^2$ . The length of a side of the square is also the diameter of the circle. The radius is $5 \;\mathrm{cm}$ .

$A=\pi {r^2} =\pi(5^2)= 25\pi \approx {78.5}$

The area is $25\pi\thickapprox 78.5 \;\mathrm{cm}^2$ .

Area of a Sector

The area of a sector is simply an appropriate fractional part of the area of the circle. Suppose a sector of a circle with radius $r$ and circumference $C$ has an arc with a degree measure of $m^\circ$ and an arc length of $s\;\mathrm{units}$ .

The sector is $\frac{m}{360}$ of the circle.
The sector is also $\frac{s}{c} = \frac{s}{{2}{\pi}{r}}$ of the circle.

To find the area of the sector, just find one of these fractional parts of the area of the circle. We know that the area of the circle is $\pi r^2$ . Let $A$ be the area of the sector.

$A = \frac{m}{360}\times \pi r^2$

Also, $A = \frac{s}{c}\times\pi r^2 = \frac{s}{{2}{\pi}{r}}\times\pi r^2 =\frac{1}{2}sr.$

Area of a Sector

A circle has radius $r$ . A sector of the circle has an arc with degree measure $m^\circ$ and arc length $s\;\mathrm{units}$ .

The area of the sector is $A \;\mathrm{square\ units}$ .

$A = \frac {m}{360}\times \pi r^2 = \frac{1}{2}sr$

Example 2

Mark drew a sheet metal pattern made up of a circle with a sector cut out. The pattern is made from an arc of a circle and two perpendicular $6-\mathrm{inch}$ radii.

How much sheet metal does Mark need for the pattern?

The measure of the arc of the piece is $270^\circ$ , which is $\frac{270}{360}=\frac {3}{4}$ of the circle.

The area of the sector (pattern) is $= \frac{3}{4}\pi r^2=\frac {3}{4}\pi \times 6^2 = 27\pi \thickapprox 84.8\;\mathrm{sq\ in}$ .

Lesson Summary

We used the idea of a limit again in this lesson. That enabled us to find the area of a circle by studying polygons with more and more sides. Our approach was very similar to the one used earlier for the circumference of a circle. Once the area formula was developed, the area of a sector was a simple matter of taking the proper fractional part of the whole circle.

Summary of Formulas:

Area Formula

Let $r$ be the radius of a circle, and $A$ the area.

$A = \pi r^2$

Area of a Sector

A circle has radius $r$ . A sector of the circle has an arc with degree measure $m^\circ$ and arc length $s\;\mathrm{units}$ .

The area of the sector is $A$ square units.

$A=\frac {m}{360}\times \pi r^2 =\frac{1}{2}sr$

Points to Consider

When we talk about a limit, for example finding the limit of the areas of regular polygons, how many sides do we mean when we talk about “more and more?” As the polygons have more and more sides, what happens to the length of each side? Is a circle a polygon with an infinite number of sides? And is each “side” of a circle infinitely small? Now that’s small!

In the next lesson you’ll see where the formula comes from that gives us the areas of regular polygons. This is the formula that was used to produce the table of areas in this lesson.

Review Questions

Complete the table of radii and areas of circles. Express your answers in terms of $\pi$ .

	Radius (units)	Area (square units)
1a.	$10$	?
1b.	?	$2.25 \pi$
1c.	?	$9$
1d.	$5 \pi$	?

Prove: The area of a circle with diameter $d$ is $\frac{\pi d^2}{4}.$
A circle is inscribed in a square.

The yellow shaded area is what percent of the square?

The circumference of a circle is $300\;\mathrm{feet}$ . What is the area of the circle?
A center pivot irrigation system has a boom that is $400 \;\mathrm{m}$ long. The boom is anchored at the center pivot. It revolves around the center pivot point once every three days, irrigating the ground as it turns. How many hectares of land are irrigated each day?
$(1\;\mathrm{hectare} = 10,000 \;\mathrm{m}^2)$
Vicki is cutting out a gasket in her machine shop. She made a large circle of gasket material, then cut out and removed the two small circles. The centers of the small circles are on a diameter of the large circle. Each square of the grid is $1 \;\mathrm{square\ inch}$ .

How much gasket material will she use for the gasket?

A security system scans all points up to $100 \;\mathrm{m}$ from is base. It scans back and forth through an angle of $65^\circ$ .

How much space does the system cover?

A simplified version of the international radiation symbol is shown below.

The symbol is made from two circles and three equally spaced diameters of the large circle. The diameter of the large circle is $12\;\mathrm{inches}$ , and the diameter of the small circle is $4\;\mathrm{inches}$ . What is the total area of the symbol?

Chad has $400\;\mathrm{feet}$ of fencing. He will use it all. Which would enclose the most space, a square fence or a circular fence? Explain your answer.

Review Answers

1a. $100 \pi$ 1b. $1.5$ 1c. $\frac{3 \sqrt{\pi}}{\pi}$ 1d. $\left(25 \pi\right)^2$
$A& = \pi r^2, r = \frac{d}{2}\ A & =\pi \left( \frac{d}{2}\right)^2=\pi \left ( \frac{d^2}{4} \right )=\frac{\pi d^2}{4}$
Approximately $21.5\%$
Approximately $7166\;\mathrm{square\ feet}$
Approximately $16.7$
Approximately $87.9\;\mathrm{square\ inches}$
Approximately $5669 \;\mathrm{m}^2$
$20 \pi \approx 62.8\;\mathrm{square\ inches}$
The circular fence has a greater area.
Square:

$P & = 4S = 400, s = 100\ A & = s^2 = 100^2 = 10,000 \;\mathrm{ft}^2$

Circle:

$C & = \pi \ d = 2\pi r = 400\ r & =\frac{400}{2 \pi}\ A & ={\pi}r^2=\pi\left(\frac{400}{{2}{\pi}}\right)^2= \frac{{200}^2}{\pi}\thickapprox 12,739 \;\mathrm{ft}^2$

Regular Polygons

Learning Objectives

Recognize and use the terms involved in developing formulas for regular polygons.
Calculate the area and perimeter of a regular polygon.
Relate area and perimeter formulas for regular polygons to the limit process in prior lessons.

Introduction

You’ve probably been asking yourself, “Where did the areas and perimeters of regular polygons in earlier lessons come from?” Or maybe not! You might be confident that the information presented then was accurate. In either case, in this lesson we’ll fill in the missing link. We’ll derive formulas for the perimeter and area of any regular polygon.

You already know how to find areas and perimeters of some figures—triangles, rectangles, etc. Not surprisingly, the new formulas in this lesson will build on those basic figures—in particular, the triangle. Note too that we will find an outstanding application of trigonometric functions in this lesson.

Parts and Terms for Regular Polygons

Let’s start with some background on regular polygons.

Here is a general regular polygon with $n$ sides; some of its sides are shown.

In the diagram, here is what each variable represents.

$s$ is the length of each side of the polygon.
$r$ is the length of a “radius” of the polygon, which is a segment from a vertex of the polygon to the center.
$x$ is the length of one-half of a side of the polygon $(2x = s).$
$a$ is the length of a segment called the apothem—a segment from the center to a side of the polygon, perpendicular to the side. (Notice that $a$ is the altitude of each of the triangles formed by two radii and a side.)

The angle between two consecutive radii measures $\frac{360^\circ}{n}$ because $n$ congruent central angles are formed by the radii from the center to each of the $n$ vertices of the polygon. An apothem divides each of these central angles into two congruent halves; each of these half angles measures $\frac{1}{2}\times\frac{360^\circ}{n}=\frac{360^\circ}{2n}=\frac{180^\circ}{n}$ .

Using Trigonometry with the Regular Polygon

Recall that in a right triangle:

$\text{sine of an angle} = \frac{\text{opposite side}}{\text{hypotenuse}} \ \text{cosine of an angle} = \frac{\text{adjacent side}}{\text{hypotenuse}}$

In the diagram above, for the half angles mentioned,

$x$ is the length of the opposite side
$a$ is the length of the adjacent side
$r$ is the length of the hypotenuse

Now we can put these facts together:

$\sin\frac{180^\circ}{n}=\frac{\mathrm{opposite\ side}}{\mathrm{hypotenuse}}=\frac{x}{r}$
$x=r\sin \frac{180^\circ}{n}$
$\cos\frac{180^\circ}{n}=\frac{\mathrm{adjacent\ side}}{\mathrm{hypotenuse}}=\frac{a}{r}$
$a=r\cos \frac{180^\circ}{n}$

Perimeter of a Regular Polygon

We continue with the regular polygon diagrammed above. Let $P$ be the perimeter. In simplest terms,

$P = ns$

Here is an alternate version of the perimeter formula.

$P & = ns = n(2x)=2nx \ P& =2nr\sin \frac{180^\circ}{n}$

Perimeter of a regular polygon with $n$ sides and a radius $r \;\mathrm{units}$ long:

$P=2nr\sin \frac{180^\circ}{n}$

One more version of the perimeter formula applies when the polygon is inscribed in a “unit circle,” which is a circle with a radius of $1$ .

$P=2nr\sin\frac{180^\circ}{n}=2n(1)\sin\frac{180^\circ}{n}=2n\sin\frac{180^\circ}{n}$

Perimeter of a regular polygon with $n$ sides inscribed in a unit circle:

$P=2n\sin \frac{180^\circ}{n}$

Example 1

A square has a radius of $6 \;\mathrm{inches}$ . What is the perimeter of the square?

Use $P=2nr\sin \frac{180^\circ}{n}$ , with $n = 4$ and $r = 6$ .

$P=2nr \sin \frac{180^\circ}{n}=2(4)(6) \sin 45^\circ=48 \left( \frac {\sqrt{2}}{2} \right) \approx 33.9\;\text{ in}.$

Notice that a side and two radii make a right triangle.

The legs are $6$ inches long, and the hypotenuse, which is a side of the square, is $6\sqrt{2}$ inches long.

Use $P = ns.$

$P=ns=4(6\sqrt{2})=24\sqrt{2} \approx33.9\;\mathrm{inches}.$

The purpose of this example is not to calculate the perimeter, but to verify that the formulas developed above “work.”

Area of a Regular Polygon

The next logical step is to complete our study of regular polygons by developing area formulas.

Take another look at the regular polygon figure above. Here’s how we can find its area, $A$ .

Two radii and a side make a triangle with base $s$ and altitude $a$ .

There are $n$ of these triangles.
The area of each triangle is $\frac{1}{2}bh=\frac{1}{2}sa$ .

The entire area is $A=n \left(\frac{1}{2}sa\right)=\frac{1}{2}(ns)a=\frac{1}{2}Pa$ .

Area of a regular polygon with apothem $a$ :

$A=\frac{1}{2}Pa$

We can use trigonometric functions to produce a different version of the area formula.

$A=\frac{1}{2}Pa=\frac{1}{2}(ns)a=\frac{1}{2}n(2x)a=nxa$ (remember that $s = 2x$ )

$A=n \left ( r\sin \frac{180^\circ}{n} \right )\left ( r\cos \frac{180^\circ}{n}\right)$ (remember that $x=r\sin\frac{180^\circ}{n}$ and $a=r\cos \frac{180^\circ}{n}$ )

$A=nr^2 \sin \frac{180^\circ}{n} \cos \frac{180^\circ}{n}$

Area of a regular polygon with $n$ sides and radius $r$ :

$A=nr^2 \sin \frac{180^\circ}{n} \cos \frac{180^\circ}{n}$

One more version of the area formula applies when the polygon is inscribed in a unit circle.

$A=nr^2 \sin \frac{180^\circ}{n} \cos \frac{180^\circ}{n}=n(1^2)\sin \frac{180^\circ}{n} \cos \frac{180^\circ}{n}=n \sin \frac{180^\circ}{n} \cos\frac{180^\circ}{n}$

(remember that $r = 1$ )

Area of a regular polygon with $n$ sides inscribed in a unit circle:

$A=n \sin \frac{180^\circ}{n} \cos \frac{180^\circ}{n}$

Example 2

A square is inscribed in a unit circle. What is the area of the square?

Use $A=n \sin \frac{180^\circ}{n} \cos \frac{180^\circ}{n}$ with $n=4$ .

$A=n \sin \frac{180^\circ}{n} \cos \frac{180^\circ}{n}=4 \sin \frac{180^\circ}{n} \cos \frac{180^\circ}{n}= 4\sin 45 \cos 45=4(0.5)=2$

The square is a rhombus with diagonals $2\;\mathrm{units}$ long. Use the area formula for a rhombus.

$A=\frac{1}{2}{d_1d_2}=\frac{1}{2}(2)(2)=\frac{1}{2}\times 4=2$

Comments: As in example 1, the purpose of this example is to show that the new area formulas do work. We can confirm that the area formula gives a correct answer because we have another way to confirm that the area is correct.

Lesson Summary

The lesson can be summarized with a review of the formulas we derived.

	Perimeter	Area
Any regular polygon	$P = ns$	$A=\frac{1}{2}Pa$
Any regular polygon	$P=2nr\sin \frac{180^\circ}{n}$	$A=nr^2 \sin \frac{180^\circ}{n} \cos \frac{180^\circ}{n}$
Regular polygon inscribed in a unit circle	$P=2n\sin \frac{180^\circ}{n}$	$A=n \sin \frac{180^\circ}{n} \cos \frac{180^\circ}{n}$

Points to Consider

We used the concept of a limit in an earlier lesson. In the Lesson Exercises, you will have an opportunity to use the formulas from this lesson to “confirm” the circumference and area formulas for a circle, which is the “ultimate” regular polygon (with many, many sides that are very short).

Review Questions

Each side of a regular hexagon is $5\;\mathrm{inches}$ long.

What is the radius of the hexagon?
What is the perimeter of the hexagon?
What is the area of the hexagon?

A regular $50-\mathrm{gon}$ and a regular $100-\mathrm{gon}$ are inscribed in a circle with a radius of $10\;\mathrm{centimeters}$ .

Which polygon has the greater perimeter?
How much greater is the perimeter?
Which polygon has the greater area?
How much greater is the area?
A regular $n-\mathrm{gon}$ is inscribed in a unit circle. The area of the $n-\mathrm{gon}$ , rounded to the nearest hundredth, is $3.14$ . What is the smallest possible value of $n$ ?

Review Answers

$5\;\mathrm{inches}$
$30\;\mathrm{inches}$
$65.0\;\mathrm{square\ inches}$
The $100-\mathrm{gon}$
$0.031\;\mathrm{cm}$
The $100-\mathrm{gon}$
$0.62\;\mathrm{cm}^2$
$56$

Geometric Probability

Learning Objectives

Identify favorable outcomes and total outcomes.
Express geometric situations in probability terms.
Interpret probabilities in terms of lengths and areas.

Introduction

You’ve probably studied probability before now (pun intended). We’ll start this lesson by reviewing the basic concepts of probability.

Once we’ve reviewed the basic ideas of probability, we’ll extend them to situations that are represented in geometric settings. We focus on probabilities that can be calculated based on lengths and areas. The formulas you learned in earlier lessons will be very useful in figuring these geometric probabilities.

Basic Probability

Probability is a way to assign specific numbers to how likely, or unlikely, an event is. We need to know two things:

the total number of possible outcomes for an event. Let’s call this $t$ .
the number of “favorable” outcomes for the event. Let’s call this $f$ .

The probability of the event, call it $P$ , is the ratio of the number of favorable outcomes to the total number of outcomes.

Definition of Probability

$P=\frac{f}{t}$

Example 1

Nabeel’s company has $12$ holidays each year. Holidays are always on weekdays (not weekends). This year there are $260$ weekdays. What is the probability that any weekday is a holiday?

There are $260$ weekdays in all.

$t=260$

$12$ of the weekdays are holidays

$f & =12 \ P & =\frac{f}{t}=\frac{12}{260} \approx 0.05$

Comments: Probabilities are often expressed as fractions, decimals, and percents. Nabeel can say that there is a $5\%$ chance of any weekday being a holiday. Note that this is (unfortunately?) a relatively low probability.

Example 2

Charmane has four coins in a jar: two nickels, a dime, and a quarter. She mixes them well. Charmane takes out two of the coins without looking. What is the probability that the coins she takes have a total value of more than $\$0.25$ ?

$t$ in this problem is the total number of two-coin combinations. We can just list them all. To make it easy to keep track, use these codes: $N1$ (one of the nickels), $N2$ (the other nickel), $D$ (the dime), and $Q$ (the quarter).

Two-coin combinations:

$N1,N2 && N1,D && N1,Q && N2,D && N2,Q && D, Q$

There are six two-coin combinations.

$t = 6$

Of the six two-coin combinations, three have a total value of more than $\$0.25$ . They are:

$N1,Q (\$0.30) && N2,Q (\$0.30) && D,Q (\$0.35)$

$f = 3$

The probability that the two coins will have a total value of more than $\$0.25$ is $P=\frac{f}{t}=\frac{3}{6}=\frac{1}{2}=0.5=50\%$ .

The probability is usually written as $\frac{1}{2},0.5$ , or $50\%$ . Sometimes this is expressed as “a $50-50$ chance” because the probability of success and of failure are both $50\%$ .

Geometric Probability

The values of $t$ and $f$ that determine a probability can be lengths and areas.

Example 3

Sean needs to drill a hole in a wall that is $14 \;\mathrm{feet}$ wide and $8 \;\mathrm{feet}$ high. There is a $2-\mathrm{foot}-$ by $-3-\mathrm{foot}$ rectangular mirror on the other side of the wall so that Sean can’t see the mirror. If Sean drills at a random location on the wall, what is the probability that he will hit the mirror?

The area of the wall is $14\times 8=112 \;\mathrm{square\ feet}$ . This is $t$ .

The area of the mirror is $2\times 3=6 \;\mathrm{square\ feet}$ . This is $f$ .

The probability is $P=\frac{6}{112} \approx 0.05$ .

Example 4

Ella repairs an electric power line that runs from Acton to Dayton through Barton and Canton. The distances in miles between these towns are as follows.

$\mathrm{Barton\ to\ Canton} = 8\;\mathrm{miles}.$
$\mathrm{Acton\ to\ Canton} = 12\;\mathrm{miles}.$
$\mathrm{Canton\ to\ Dayton} = 2\;\mathrm{miles}.$

If a break in the power line happens, what is the probability that the break is between Barton and Dayton?

Approximately $71\%$ .

$t & = \text{the distance from Acton to Dayton} = 4 + 8 + 2 = 14\ \text{miles}. \ f & = \text{the distance from Barton to Dayton} = 8 + 2 = 10\ \text{miles}. \ P & =\frac{f}{t}=\frac{10}{14}=\frac{5}{7} \approx 0.71=71\%$

Lesson Summary

Probability is a way to measure how likely or unlikely an event is. In this section we saw how to use lengths and areas as models for probability questions. The basic probability ideas are the same as in non-geometry applications, with probability defined as:

$\text{Probability}=\frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$

Points to Consider

Some events are more likely, and some are less likely. No event has a negative probability! Can you think of an event with an extremely low, or an extremely high, probability? What are the ultimate extremes—the greatest and the least values possible for a probability? In ordinary language these are called “impossible” (least possible probability) and “certain” or a “sure thing” (greatest possible probability).

The study of probability originated in the seventeenth century as mathematicians analyzed games of chance.

For Further Reading

French mathematicians Pierre de Fermat and Blaise Pascal are credited as the “inventors” of mathematical probability. The reference below is an easy introduction to their ideas. http://mathforum.org/isaac/problems/prob1.html

Review Questions

Rita is retired. For her, every day is a holiday. What is the probability that tomorrow is a holiday for Rita?
Chaz is “on call” any time, any day. He never has a holiday. What is the probability that tomorrow is a holiday for Chaz?
The only things on Ray’s refrigerator door are green magnets and yellow magnets. Ray takes one magnet off without looking.
1. What is the probability that the magnet is green?
2. What is the probability that the magnet is yellow?
3. What is the probability that the magnet is purple?
Ray takes off two magnets without looking.
1. What is the probability that both magnets are green?
2. What is the probability that Ray takes off one green and one yellow magnet?
Reed uses the diagram below as a model of a highway.
1. What is the probability that the accident is not between Canton and Dayton?
2. What is the probability that the accident is closer to Canton than it is to Barton?

Reed got a call about an accident at an unknown location between Acton and Dayton.

A tire has an outer diameter of $26\;\mathrm{inches}$ . Nina noticed a weak spot on the tire. She marked the weak spot with chalk. The chalk mark is $4\;\mathrm{inches}$ along the outer edge of the tire. What is the probability that part of the weak spot is in contact with the ground at any time?
Mike set up a rectangular landing zone that measures $200\;\mathrm{feet}$ by $500\;\mathrm{feet}$ . He marked a circular helicopter pad that measured $50$ feet across at its widest in the landing zone. As a test, Mike dropped a package that landed in the landing zone. What is the probability that the package landed outside the helicopter pad?
Fareed made a target for a game. The target is a $4-$ foot-by $-4$ -foot square. To win a player must hit a smaller square in the center of the target. If the probability that players who hit the target win is $20\%$ , what is the length of a side of the smaller square?
Amazonia set off on a quest. She followed the paths shown by the arrows in the map.

Every time a path splits, Amazonia takes a new path at random. What is the probability that she ends up in the cave?

Review Answers

$1, 100\%$ , or equivalent
$0$
1. $\frac{2}{5}, 0.4, 40\%$ , or equivalent
2. $\frac{3}{5}, 0.6, 60\%$ , or equivalent
3. $0$
4. $\frac{2}{15} \approx 0.13$ or equivalent
5. $\frac{4}{15} \approx 0.27$ or equivalent
1. $\frac{12}{14}=\frac{6}{7} \approx 0.86=86\%$
2. $\frac{6}{14}=\frac{3}{7} \approx 0.43=43\%$
Approximately $0.05=5\%$
Approximately $98\%$
Approximately $1.78\;\mathrm{feet}$
$\frac{5}{12} \approx 0.42$ or equivalent