RATIOS AND PROPORTIONS Ratios are a comparison of two quantities. Proportions are math statements composed of two equivalent ratios. Proportions can be very helpful when applied to solving practical real-world problems. This unit begins with simplifying ratios and solving basic proportions. Ratios and proportions are then applied in problems about similar figures, unit rates, and scale drawings. The unit concludes with an exploration of Venn diagrams. Venn diagrams are diagrams that are used to examine similarities and differences of groups of objects or numbers.

Ratios and Proportions

Rates

Similar Triangles

Similar Polygons

Scale Drawings

Venn Diagrams

Ratios and Proportions Ratio Models A ratio is a comparison of two quantities. The colors of these leaves may be compared many ways.

There are three ways to write ratios. To compare the yellow leaves to green leaves, you can write:

44 to 3 4:33

or or

The ratio of yellow leaves to green leaves is 4 : 3.

Example 1: Write the ratio of red leaves to brown leaves three ways.

22 to 3 2:33

or or

The ratio of red leaves to brown leaves is 2 : 3. Example 2: Write the ratio of green leaves to all of the leaves as a fraction in simplest form.

33 to 12 3:1212

or or

The ratio of green leaves to all of the leaves is 3 : 12. This ratio can be simplified; so, use the fraction form of the ratio and reduce.

3 3 112 3 4

÷ =

The ratio of green leaves to all of the leaves is 1 : 4. This means that one in four leaves is green.

Example 3: Write the ratio of all of the leaves to the red leaves as a fraction in simplest form.

1212 to 2 12:22

or or

The ratio of all of the leaves to the red leaves is 12 to 2. This ratio can be simplified; so, use the fraction form of the ratio and reduce.

12 2 62 2 1÷ =

The ratio of all of the leaves to the red leaves is 6 : 1. This means for every six leaves in the group, one is red.

*Ratios such as 6/1 or 5/3 remain as improper fractions when written in fraction form since a ratio is a comparison of two quantities.

Proportions

Using ratios to set up a proportion is a very useful problem-solving technique that can be used in a variety of everyday problems.

Example 4: Cindy surveyed her home-room class and found that 11 out of 25 students played video games at home at least once every day of the school week. There are 500 students in the ninth grade at her school. Based on Cindy’s survey, predict how many ninth grade students in the school play video games at least once every day of the school week.

To solve, first find the comparison ratio.

11 (play a video game at least once every day of the school week)25 (all students surveyed)

=

Then, set up a proportion where both numerators represent the same idea and both denominators represent the same idea.

Thinking through the problem with brief words can be helpful. 11 out of 25 play video games…how many out of 500 play video games…?

The proportion is 11 25 500

n= .

*The cross products of a proportion are equivalent. Use this technique to solve for n.

11500

11 500

Cross Multiply

Simplfy

25 5500

25

25

Divide

220

n

n

n

n

=

=

=

=

× ×

Cindy estimated that, based on her survey, about 220 ninth grade students play a video game at least once every day of the school week.

Rates rate – A rate is a ratio that is a comparison of two quantities that are different kinds of units. unit rate – A unit rate is a ratio that is a comparison of a quantity to one. (The rate is expressed with a denominator of one.)

Sample Rates:

miles per hour (mph) miles per gallon (mpg) revolutions per minute (rpm)

Example 1: Find the average unit rate of speed for a car traveling 700 miles in 12 hours. Round the answer to the nearest tenth.

Set up a proportion comparing miles to hours on both sides of the proportion.

The rate of speed is approximately 58.3 miles per hour.

miles 700hours 12 1

n=

Cross multiply.

12n = 700

Divide by 12 to find n.

n ≈58.3 mph

To solve for a unit rate, solve for ONE.

Unit Prices and Comparative Shopping unit price - The unit price for an item is the cost for one unit of measure. The unit price may be the cost per ounce, per pound, per item, or other units. When you shop in a local store, you can use unit prices to determine the better buy. To find the unit price of any item, divide the price by the number of units it contains.

Example 2: Alex noticed that the price of two jars of pickles on the grocery store shelf has the following labels: He decided to compare the two brands to determine the better buy.

First:

Change 1 pound 2 ounces to ounces. 1 pound + 2 ounces (1 pound = 16 ounces) 16 ounces + 2 ounces = 18 ounces

Brand A

Pickles

1 lb 2 oz @ $1.29

Brand B

Pickles

15 oz @ $0.89

Next:

Compute the unit price per ounce for each brand.

price # of units unit price

1.29 18 0.07166... $.07

0.89 15 0.05933... $.06

÷ =

÷ = ≈

÷ = ≈

Brand A

Brand B

Brand B is the better buy because the unit price is lower.

Similar Triangles Similar triangles are triangles that have the same shape. That is, the corresponding sides of the triangles are proportional and the corresponding angles are congruent (same size). Let’s examine each of these properties separately.

If two triangles are similar, then their corresponding angles are congruent.

*The math symbol to denote similarity is “~”. *The math symbol to denote congruency is “≅ ”. * DEF ~ TUV is read “Triangle DEF is similar to triangle TUV.”

Example 1: DEF ~ TUV. State the corresponding congruent angles, and then find the measures of angles E and V.

*The red square in each triangle denotes a right angle. Angle D and angle T are right angles.

The corresponding congruent angles are:

D T E U F V≅ ∠ ∠ ≅∠ ∠ ≅∠

D T is read "angle D is congruent to angle T."*∠ ≅∠

E

D F

U

T V

65°

25°

Angle D and Angle T

Angle D and angle T are right angles; therefore, they have the same measure.

We can state: D Tm m∠ = ∠

This statement is read, “The measurement of angle D equals the measurement of angle T.” Since we know that right angles measure 90 degrees, we can state:

D 90 T 90m m∠ = ° ∠ = ° We could also simply state: D T 90m m∠ = ∠ = °

Angle E and Angle U

Angle E and angle U are corresponding angles; thus, they are congruent and they have equal measures.

E U, therefore E U

Given, U 65 Therefore, E 65

m m

m m

∠ ≅ ∠ ∠ = ∠

∠ = ° ∠ = °

E

D F

U

T V

65°

25°

Angle F and Angle V

Angle F and angle V are corresponding angles; thus, they are congruent and they have equal measures.

F V, therefore F V

Given, F 25 Therefore, V 25

m m

m m

∠ ≅ ∠ ∠ = ∠

∠ = ° ∠ = °

If two triangles are similar, then their corresponding sides are proportional.

Example 2: ABC ~ XYZ. Find the length of segment AC that is labeled with x given the measurements shown in the figure.

Since the triangles are similar triangles, the corresponding sides are proportional and there exists a common ratio between the pairs of corresponding sides. This common ratio is called the scale factor and is the same between any pair of the corresponding sides in the two given similar triangles.

E

D F

U

T V

65°

25°

ABC ~ XYZ

x

B

A C

Y

X Z

16 18 8 9

4

Therefore, the corresponding sides can be written as equivalent ratios.

16 188 9 4

AB BC ACXY YZ XZ

x

= =

= =

Method 1: To solve for x, set up a proportion based on two of the ratios.

Let’s use:

16 Substitute the given values.8 4

8 64 Cross multiply.

8 Divide.

AB ACXY XZ

x

x

x

=

=

=

=

The length of the missing side (x) is 8 units.

Now, we’ll use a second proportion to check the answer.

18 Substitute the given values.9 4

9 72 Cross multiply.

8 Divide.

BC ACYZ XZ

x

x

x

=

=

=

=

Method 2: A second way to solve for the missing side is to determine the scale factor. Examine the given corresponding ratios and simplify.

16 182 28 9

AB BCXY YZ

= = = =

Both of these ratios simplify to 2. The scale factor is 2.

4AC xXZ

=

Therefore,

24

(4)(2)8

x

xx

=

==

The length of the missing side (x) is 8 units.

Indirect Measurements Take a look at two scenarios where similar triangles are applied to find lengths and distances indirectly.

Scenario 1: Bill is preparing to cut down a tree in the yard. He doesn’t want to actually fell the tree until he has a pretty good estimate of the height. Then he notices that both he and the tree are casting a shadow. His shadow is 3.5 feet long, while the shadow of the tree is 19 feet long. If Bill is six feet tall, how tall is the tree?

Since similar triangles are formed by the tree casting a shadow and Bill casting a shadow, the height of the tree can be determined indirectly. Use the values from the similar triangles to set up a proportion to solve for the height (h) of the tree.

height of tree shadow of treeBill's height Bill's shadow

196 3.5h

=

=

Cross multiply and solve.

3.5 19 63.5 114

33 (rounded to nearest whole number)

hh

h

= ×=

=

The height of the tree is approximately 33 feet.

h

19 ft

6 ft

3.5 ft

Now, let’s take a look at the problem of finding the distance across a lake which is difficult to measure under normal circumstances. Measurements can be made across dry land and the angles can be measured with a surveyor tool. If enough information is gathered, the distance across the lake can be determined indirectly.

Scenario 2: Use the measurements shown below and similar triangles to find the distance (d) across the lake.

*Notice that the triangles are similar, but turned in different directions. The corresponding sides are denoted by different colors.

4 meters 10 meters 12 meters

d

Now, set up a proportion of values to solve for the distance (d). We will use the two legs of the right triangles, not the hypotenuses, to write a proportion in words first, and then solve. shorter leg of large right longer leg of large right shorter leg of small right longer leg of small right

124 10

d

=

=

Cross multiply and solve.

4 12 104 120

1204

30

dd

d

d

= ×=

=

=

The distance across the lake is approximately 30 meters.

Use these ideas to determine proportions using indirect measurement that will give solutions geometrically that cannot easily be done otherwise.

21

18 10.5

15

I

L K

J

5

6

7

3.5 P O

N M

Similar Polygons similar figures – Similar figures are figures that have the same shape, but are different in size. similar polygons – Similar polygons are polygons that have congruent corresponding angles and the measures of their corresponding sides are proportional. *The math symbol to denote similarity is “~”. *The math symbol for congruency is “≅ ”.

Example 1: Quadrilateral IJKL and MNOP are similar. (a) State the congruent angles. (b) Write the corresponding sides as proportional ratios. Quadrilateral IJKL ~ Quadrilateral MNOP

*The corresponding congruent angles are marked with curved “hash marks”.

a.) The corresponding congruent angles are:

denoted by one curved hash mark in the figures

denoted by two curved hash marks in the figures

denoted by three curved hash marks in the figures

denoted by four curved hash marks in t

I M

J N

K O

L P

∠ ≅ ∠

∠ ≅ ∠

∠ ≅ ∠

∠ ≅ ∠ he figures

b.) The corresponding proportional sides are:

21 18 15 10.5 7 6 5

33 1.5

IJ JK LK ILMN NO PO MP

= = =

= = = =

All the ratios reduce to the same scale factor, 3/1.

scale factor – The scale factor for two similar polygons is the ratio of the lengths of any two corresponding sides.

The scale factor of quadrilateral IJKL to quadrilateral

MNOP is 31

or 3.

When comparing the other way around, the scale factor of

quadrilateral MNOP to quadrilateral IJKL is 13

.

Refer to the figure below to solve the next three examples.

21 77 7

18 66 6

15 55 5

10.5 3.53.5 3.

31

31

31

315

⎡ ⎤÷ =⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

÷ =⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥÷ =⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥÷ =⎢ ⎥⎣ ⎦

Given: Quadrilateral ABCD ~ Quadrilateral EFGH

Example 1: Find the scale factor. Use a pair of corresponding sides that are given. For this problem, use sides BC and FG to write the ratio and simplify.

27 9 9The scale factor is .15 5 5

BCFG

= =

27

D

y + 5

C

A

B x

15

H

20 G

E

F1

Example 2: Find the value of x. Write a proportion using the values of corresponding sides that are given and the variable x. Sides BC and FG are corresponding sides and the values are given. The side (EF) that corresponds with x is also given.

18 9: which is the scale factor.10 5

Check =

27 10 15

(15) 10(27) 15 270 18

AB BCEF FG

x

xxx

=

=

===

27

D

y + 5

C

A

B x

15

H

20 G

E

F 10

Example 3: Find the value of y. Write a proportion using the values of corresponding sides that are given and the variable y. Sides BC and FG are corresponding sides and the values are given. The side (EH) corresponding with y is also given.

31 5 36 9: which is the scale factor.20 20 5

Check += =

5 27 20 15

( 5)(15) 20(27) 15 75 540

15 465 31

AD BCEH FG

y

yy

yy

=

+=

+ =+ =

==

27

D

y + 5

C

A

B x

15

H

20 G

E

F 10

Scale Drawings A scale drawing or a scale model is used to represent an object that is too large or too small to be drawn or built at actual size. Examples are blueprints, maps, models of vehicles, and models of animal anatomy. A scale is determined by the ratio of a given length in a drawing or a model to its corresponding actual length. The blueprint for a house is given below superimposed on a grid.

In order to determine the scale, you can complete the following. How many units wide is the largest bedroom including the bath? 8 units The actual width of the master bedroom is 25 feet. Write a ratio comparing the drawing width to the actual width. 8 units : 25 feet

Garage

W D

ClosetCloset

Bedroom

Bedroom Bedroom

Living Room

Kitchen

Bath

Bath

Deck25 ft

= 3.125 ft

Simplify the ratio above and compare it to the scale shown at the bottom of the drawing.

8 units : 25 units means 8 to 25

8 Write 8 to 25 as a fraction.25

8 8 Reduce by 8 so that the numerator is one.25 8

1 Simplify3.125

1: 3.125 Write the fraction as a ratio.

÷÷

According to the blueprint, the length of one unit equals 3.125 feet. Therefore, the ratio and the scale indicate that the length of one unit on the drawing is equal to 3.125 feet in reality. Distances on a scale drawing are proportional to distances in real-life. *Note: A general rule for scale models is that scales and scale factors are written so that the drawing length occurs first in the ratio.

Map Distances The following map was generated at http://maps.yahoo.com. Point A represents Quaker City, OH. Point B represents Mason, OH. On paper mark the length of one bar in the key that represents 50 miles. Then, hold the paper along the highlighted path on the map from Quaker City, OH, to Mason, OH, and count how many lengths of the bar there are between the two communitites traveling along the marked route.

It takes about 3 of plus another 34

of to get from point

A to point B (3.75).

= 50 miles Key:

Let x represent the actual distance to point B, write a proportion, and then solve.

1 3.75 Map DistanceProportion Comparing: 50 Actual Distance

(1) 50(3.75) Cross Multiply

187.5 Simplify

x

x

x

=

=

=

The actual distance from point A to point B is about 187.5 miles.

Venn Diagrams There are 50 students in Lakewood High Marching Band. The students who play brass and/or woodwind instruments were asked what instrument or instruments he or she plays. The results were recorded in the table shown below. Students who play a percussion instrument only were not surveyed; thus, they did not respond to the survey. How many students play a percussion instrument?

What is given?

Sis students play both brass and woodwind instruments. Twenty-seven students play brass instruments. Twenty-five students play woodwind instruments. The total number of students in the band is 50.

What is asked?

How many students play a percussion instrument?

Results of Band Survey Play brass 27 Play woodwind 25 Play both 6

A Venn diagram will help clarify this problem. The rectangle represents the total number of students in the band. The circle on the left (yellow circle) represents the number of students that play brass instruments. The circle on the right (blue circle) represents the number of students that play woodwind instruments. The green overlapping area of the two circles represents the number of students who play both brass and woodwind instruments. The students who play percussion will be represented outside the circles.

First, determine the number of students who play ONLY brass instruments. Subtract 6 (play both) from the 27 who responded with a “yes” for brass.

27 – 6 = 21

There are 21 students who play ONLY a brass instrument.

50 Students in Band

6 Brass Woodwind

50 Students in Band

Brass 21

Woodwind 6

*Note: The total in the yellow circle (brass) must equal 27 students, including the part of the yellow circle that overlaps with the blue circle. Thus, 6 students play brass and woodwind, and 21 play brass instruments only. Next, determine the number of students who play ONLY woodwind instruments. Subtract 6 (play both) from the 25 who responded with a “yes” for woodwind.

25 – 6 = 19

There are 19 students who play ONLY a woodwind instrument.

*Note: The total in the blue circle (woodwind) must equal 25 students, including the part of the blue circle that overlaps with the yellow circle. Thus, 6 students play brass and woodwind, and 19 play woodwind instruments only. Now, determine the total number of students who participated in the survey (play brass and/or woodwind) and how many who did not (play percussion). Look at the circles in the chart. All of the students who participated in the survey are included in the circles.

21 + 6 + 19 = 46

50 Students in Band

Brass 21

Woodwind 19

6

Therefore, the number of students who do NOT play a brass instrument or a woodwind instrument can be determined by subtracting 46 from 50. These are the students who play the percussion instruments.

50 – 46 = 4 The Venn diagram below is now complete. The students who play either a brass instrument and/or a woodwind instrument are represented inside the circles and the students who play a percussion instrument are represented outside the circles.

The Venn diagram is a tool to help depict logic problems that involve overlapping groups. To check the solution, add all parts of the Venn diagram to see if the total is 50.

21 + 6 + 19 + 4 = 50

50 Students in Band

Brass 21

4 (Percussion)

Woodwind 19

6