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## High school geometry

### Course: High school geometry>Unit 4

Lesson 1: Definitions of similarity

# Getting ready for similarity

Practicing identifying proportional relationships and solving equations with proportions helps us get ready to learn about similarity.
Let’s refresh some concepts that will come in handy as you start the similarity unit of the high school geometry course. You’ll see a summary of each concept, along with a sample item, links for more practice, and some info about why you will need the concept for the unit ahead.
This article only includes concepts from earlier courses. There are also concepts within this high school geometry course that are important to understanding similarity. If you have not yet mastered the Congruent triangles lesson or the Dilations preserved properties lesson, it may be helpful for you to review those before going farther into the unit ahead.

## Identifying proportional relationships

### What is this, and why do we need it?

A relationship between two quantities is proportional if the ratio between those quantities is always equivalent. We will look at side length ratios to find out whether triangles are similar or not.

### Practice

Problem 1
Triangle A has a height of 2, point, 5, start text, space, c, m, end text and a base of 1, point, 6, start text, space, c, m, end text. The height and base of triangle B are proportional to the height and base of triangle A.
Which of the following could be the height and base of triangle B?
Choose 3 answers:

For more practice, go to Proportional relationships.

### Where will we use this?

Here are a few of the exercises where reviewing proportional relationships might be helpful:

## Solve equations with proportions

### What is this, and why do we need it?

When two ratios are equal, we create a proportion equation. If we multiply the equation by both denominators, we can solve the resulting equation just like a linear (or quadratic, but not in this unit) equation. We will set up equations with proportions to find lengths in similar figures.

### Practice

Problem 2.1
• Current
Solve for m.
Do not round. If needed, write your answer as a fraction.
start fraction, 8, divided by, 10, end fraction, equals, start fraction, 6, divided by, m, end fraction
m, equals

For more practice, go to Solving proportions 2.

### Where will we use this?

Here are a few exercises where reviewing proportions equations might be helpful.

## Want to join the conversation?

• i do not understand why he give me a question about:q/q+5=12/27.and suddenly next stop just became:27q=12q+60.what the hell is that number 60 coming form?
(⊙⊙?)
(15 votes)
• When you cross multiply, you have to make sure to distribute if a numerator/denominator has multiple terms in it. Here's how the equation looks directly after you cross-multiply:
(27) q = (q + 5) 12
The 12 distributes to both the q and the 5, creating:
27q = 12q + 60
15q = 60
q = 4
(29 votes)
• how do you math?
(12 votes)
• ... the q/q+5=12/27 is 4. It's not that hard tbh
(3 votes)
• bro no ones cares lmao
(12 votes)
• Does anyone know practical applications of similarity concepts in common engineering fields? I'm interested in the practical applications of these otherwise abstract concepts, and was wondering if anyone had any interesting insights. Thanks!
(5 votes)
• Say an engineer makes a prototype of a design. A company likes it but wants to adjust the size a bit. The engineer must use similarity to 'scale up' or 'scale down' the prototype by a certain amount.
(7 votes)
• why did you put 60 over 8?
(3 votes)
• 8/10 = 6/x
you cross multiply
8 * x = 8x
6 * 10 = 60
then you dive 60 over 8
which equals 7.5
(10 votes)
• why is it 10m and how does that get rid of the ten when its not negative
(5 votes)
• It is a fraction...
(0 votes)
• why is it 10m and how does that get rid of the ten when its not negative
(2 votes)
• You multiply the 10 by 1/10 so you have 10/10 which simplifies into 1
(3 votes)
• why is this important in life.
(1 vote)
• its not tbh
(2 votes)
• For the last problem, I solved it by looking at the percentage of 12/27 and then realized that 4/9 has the same rate. Is there a more efficient way of doing this?
(1 vote)
• Reduce the fraction, solve for q; in this case, solving for q becomes unnecessary once you've reduced, because most people will quickly recognize that q must be four.

This actually brings up a good general principle in math: do what you know you can do, then look at the problem again, and see if it has gotten easier. Simplifying, expanding, factoring, substituting equivalent expressions, are all good things to try if you get stuck. Of course, you'll want a pretty good idea of how to do these already when you need them, say on a test, so do a variety of exercises to expose yourself to different types of problems.
(2 votes)