Main content

### Course: High school geometry > Unit 8

Lesson 1: Circle basics# Getting ready for circles

Everything we've learned about angle relationships and proportions in other figures also applies in figures with circles and parts of circles.

Let’s refresh some concepts that will come in handy as you start the circles 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 circles. If you have not yet mastered the Definitions of similarity lesson, it may be helpful for you to review that before going farther into the unit ahead.

## Circumference and area of parts of circles

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

What's the area of a semicircle (a half circle)? It's half the area of the full circle. What's the arc length of $\frac{1}{3}$ of a circle? It's $\frac{1}{3}$ of the circumference of the full circle. In high school geometry, we'll generalize from these common fractions to be able to find the arc length and area for parts of circles given the radius and any central angle measure.

### Practice

For more practice, go to Circumference of parts of circles and Area of parts of circles.

### Where will we use this?

Here are a few of the exercises where reviewing the circumference and area of parts of circles might be helpful:

## Solving proportions

### 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. The ratio between the area of a sector and the area of the whole circle is equal to the ratio between the central angle measure of the sector and the central angle measure of the whole circle. The same is true for the ratio between the arc length of a sector and the circumference of the whole circle.

### Practice

For more practice, go to Solving proportions.

### Where will we use this?

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

## Simplifying complex fractions

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

A complex fraction is a fraction where the numerator, denominator, or both are also fractions. Any proportional relationship could involve fractional values, but fractions are especially common when we use radians as an angle measure.

### Practice

For more practice, go to Simplify complex fractions.

### Where will we use this?

Here are a couple of the exercises where reviewing complex fractions might be helpful:

## Using angle relationships

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

All of the angle properties when angles share a vertex or are part of the same triangle still apply when those angles are in a figure with a circle. Do the angles combine to form a straight angle? Then their measures sum to $180\mathrm{\xb0}$ . Do they combine to form a complete turn? Then their measures sum to $360\mathrm{\xb0}$ . We can find the total angle measure of inscribed and circumscribed shapes by decomposing them into triangles.

### Practice

For more practice, go to Finding angle measures between intersecting lines and Angles of a polygon.

### Where will we use this?

Here are a couple of the exercises where reviewing angle relationships might be helpful:

## Solving equations with the unknown on both sides

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

Congruent parts of figures have equal measures. When those measures both involve the unknown, we can often still solve for the value of the unknown by rewriting the equation.

### Practice

For more practice, go to Equations with variables on both sides.

### Where will we use this?

Here is an exercise where reviewing solving equations with the unknown on both sides might be helpful:

## Finding angle measures in isosceles triangles

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

The angles opposite the congruent sides of an isosceles triangle are congruent. Because all radii of a circle are congruent, triangles with those radii as sides must be isosceles. We'll use that fact to prove an important relationship between central angles and inscribed angles on the same arc.

### Practice

For more practice, go to Find angles in isosceles triangles.

### Where will we use this?

Here is an exercise where reviewing angles in isosceles triangles might be helpful:

## Want to join the conversation?

- i'm only in 8th grade but just wanted to see this and so far it looks impossible.(14 votes)
- It only looks impossible until you have the basics down. When you come back to it in high school it will look easy. Circles are one of the least complicated shapes I have ever seen.(12 votes)

- why we gotta do this(9 votes)
- onb this is annoying(6 votes)

- @ problem 4.1, I used a different but longer method to solve x, by using <aec and <deb vertical angle congruence.

<bec (160 deg) + <aec (? deg) = 180 deg

<aec = 20 deg

<aec = <deb

<deb = 20 deg

<deb (20 deg) + <fed (115 deg) = <feb (135 deg)

<aef(x) + <feb(135) = 180 deg

180 - 135 = 45

<aef (x) = 45 deg(7 votes)- That is some big brains there, nj_lemons. I don’t think I can do that and I am in tenth grade! Good job!(5 votes)

- I feel like geometry is more of critical thinking rather than Algebra 2. I am doubling up this year and I'm a freshman in highschool, but all you guys struggling, you will get this! Just revisit the basic theorems of angles and memorize the formulas. You guys can do it and happy learning!(7 votes)
- I think the second question problem 1.2's answer is wrong or my calculator is wrong, I got 84.82 my way 3/4[pi(6)^2] it said i was wrong. I then check the answer it said 27pi or 84.78 I put 27pi in a calculator and it gave me 84.82 now I am confused.(3 votes)
- Thats because its 36 * 3/4 * pi or 27 pi it might be a rounding error or something(1 vote)

- Gremio menor time do brasil(3 votes)
- how do i do all of this(1 vote)