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Course: High school geometry (staging) > Unit 9

Lesson 1: Circle basics
Math
High school geometry (staging)
Circles
Circle basics
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Getting ready for circles

Google Classroom
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 13 of a circle? It's 13 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

Problem 1.1
The fourth of a circle with a radius length of seven and an arc of an unknown length.
Find the arc length of the partial circle.
Either enter an exact answer in terms of π or use 3.14 for π and enter your answer as a decimal.
units

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

Problem 2.1
Solve for n.
11n=85
n=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

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

Problem 3.1
Which expression is equivalent to the following complex fraction?
(74)(98)
Choose 1 answer:

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°. Do they combine to form a complete turn? Then their measures sum to 360°. We can find the total angle measure of inscribed and circumscribed shapes by decomposing them into triangles.

Practice

Problem 4.1
What is the value of x in the following figure?
A point labeled E and six points around it. All six points are at different positions. Point A is located at nine o'clock. Point B is located at three o'clock. Point C is located at eight o'clock. Point D is located at two o'clock. Point F is located at eleven o'clock. Point G is located at five o'clock. Six rays point from the same endpoint, E through each of the points. Angle A E F is x degrees. Angle F E D is one hundred fifteen degrees. Angle C E B is one hundred sixty degrees.
NOTE: Angles not necessarily drawn to scale.
x=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
°

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

Problem 5
Solve for b.
7b−15=5b+17
b=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

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

Problem 6.1
Triangle PMN is isosceles because PM― and NM― are radii of circle M.
A circle that is centered around point M. Triangle M N P is inscribed in the circle. Sides M P and M N are congruent. Points P and N lie on the circumference of the circle. Angle M N P is fifty degrees. Angle N M P is x degrees. Side M P is seven units. Side M N is seven units. Side P N is approximately nine units.
What is the value of x in △PMN?
x=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
°

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:

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