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# 9.18: Unknown Measures of Similar Figures

Difficulty Level: Basic Created by: CK-12
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Have you ever been stuck figuring out a math problem?

Jessica saw this problem on her homework assignment. She knows that she will need to use proportions in some way to figure out the length of the missing side, she just isn't sure how to do it.

Do you know?

This Concept will teach you how to tackle problems like this one.

### Guidance

Once you know how to locate the corresponding sides of similar triangles, we can write ratios to compare the lengths of sides.

First, identify the corresponding sides of these two similar triangles.

$\frac{LM}{OP} = \frac{LN}{OQ} = \frac{MN}{PQ}$

Now we have been given side lengths for each pair of corresponding sides. These have been written in a proportion or a set of three equal ratios. Remember that there is a relationship between the corresponding sides because they are parts of similar triangles. The side lengths of the similar triangles form a proportion .

Let’s substitute the given measurements into our formula.

$\frac{6}{3} = \frac{8}{4} = \frac{4}{2}$

There is a pattern with the ratios of corresponding sides. You can see that the measurement of the each side of the first triangle divided by two is the measure of the corresponding side of the second triangle.

We can use patterns like this to problem solve the length of missing sides of similar triangles.

Here we have two similar triangles. One is larger than the other, but they are similar. They have the same shape but a different size. Therefore, the corresponding sides are similar.

If you look at the side lengths, you should see that there is one variable. That is the missing side length. We can figure out the missing side length by using proportions. We know that the corresponding side lengths form a proportion. Let’s write ratios that form a proportion and find the pattern to figure out the length of the missing side.

$\frac{AB}{DE} & = \frac{AC}{DF} = \frac{BC}{EF} \\\frac{5}{10} & = \frac{15}{x} = \frac{10}{20}$

Looking at this you can see the pattern. The side lengths of the second triangle are double the length of the corresponding side of the first triangle.

Using this pattern, you can see that the length of $DF$ in the second triangle will be twice the length of $AC$ . The length of $AC$ is 15.

15 $\times$ 2 $=$ 30

The length of $DF$ is 30.

Practice solving these proportions.

#### Example A

$\frac{6}{12} = \frac{x}{24} = \frac{3}{6}$

Solution: $x = 12$

#### Example B

$\frac{12}{x} = \frac{16}{4} = \frac{20}{5}$

Solution: $x = 3$

#### Example C

$\frac{8}{2} = \frac{16}{4} = \frac{x}{1}$

Solution: $x = 4$

Here is the original problem once again.

Jessica saw this problem on her homework assignment. She knows that she will need to use proportions in some way to figure out the length of the missing side, she just isn't sure how to do it.

Do you know?

Now the first thing that we can do is to set up a proportion to solve for the missing side. Remember that a proportion is two equal ratios. We can set up and compare the corresponding sides.

Here is our proportion.

$\frac{KJ}{5} = \frac{6}{4}$

Our proportion is written so that the corresponding sides form the two ratios of the proportion. We can say that $KJ$ is our unknown in this proportion.

Do you remember how to solve proportions?

We can see a clear relationship between five and four, so we need to use cross products.

$KJ \times 4 &= 4KJ\\5 \times 6 &= 30\\4KJ &= 30$

Now we can solve the equation for $KJ$ by dividing both sides of the equation by 4.

$30 \div 4 &= 7.5\\KJ &= 7.5$

The side length of $KJ$ is 7.5.

### Vocabulary

Here are the vocabulary words that are found in this Concept.

Congruent
having the same size and shape and measurement
Similar
having the same shape, but not the same size. Similar shapes are proportional to each other.
Corresponding
matching-corresponding sides between two triangles are sides that match up
Ratio
a way of comparing two quantities
Proportion
a pair of equal ratios.

### Guided Practice

Here is one for you to try on your own.

$\frac{8}{10} = \frac{4}{5} = \frac{2}{x}$

To look at the relationships between each side length, we can begin by looking for a pattern of equal ratios.

The first ratio was divided in half to equal the second ratio.

The numerator of the second ratio was divided in half to equal the numerator of the third ratio.

The denominator of the third ratio is unknown.

We can divide the denominator of the second ratio in half to equal the missing denominator.

$5 \div 2 = 2.5$

$x = 2.5$

### Video Review

Here are videos for review.

### Practice

Directions : Use the figures to answer the following questions.

1. Are these two triangles similar or congruent?

2. How do you know?

3. Which side is congruent to $AB$ ?

4. Which side is congruent to $AC$ ?

5. Which side is congruent to $RS$ ?

6. Look at the following proportion and solve for missing side length $x$ .

$\frac{7}{3.5} & = \frac{x}{3.5} = \frac{6}{y} \\x & = \underline{\;\;\;\;\;\;\;\;\;\;\;}$

7. What is the side length for $y$ ?

8. How did you figure these out?

Directions : Figure out the missing value in each pair of ratios.

9. $\frac{6}{12} = \frac{x}{24}$

10. $\frac{8}{12} = \frac{x}{3}$

11. $\frac{9}{10} = \frac{18}{y}$

12. $\frac{4}{5} = \frac{x}{2.5}$

13. $\frac{16}{20} = \frac{4}{y}$

14. $\frac{19}{21} = \frac{x}{42}$

15. $\frac{9}{54} = \frac{6}{y}$

Basic

Oct 29, 2012

Aug 18, 2014