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Parts of a circle Trigonometry SOHCAHTOA Pythagorean theoremGeometry proof

Here you will learn about chords of circles. You will learn what a chord of a circle is, theorems that involve chords, and the application of these theorems. You will explore the proof of the theorems and how to use them to solve more complex problems.

Students first learn how to work with these circles and chords in a geometry course where they spend time delving into the proofs of theorems and application of those theorems. Circle theorems are essential for success on graduation assessments as well as college entrance assessments.

The **chord of a circle** is a straight line that connects two points on the edge of the circle or circumference of a circle. The longest chord in a circle is the diameter of the circle.

As the chords of a circle get further away from the center of the circle, the shorter they become.

Looking at circle A, you can visually see how chord BF is shorter than chord CE. Line segment DG has endpoints on the edge of the circle (circumference of the circle) and goes through the center of the circle. This means that chord DG is the diameter of the circle and also the longest chord in the circle.

Letβs look at when the radius (or diameter) is perpendicular to a chord.

In circle C, \, CE is the radius and it is perpendicular to chord AB.

Notice how radius CE also bisects chord AB at a 90^{\circ} angle. Bisect means to cut in half which means that the lengths AD=BD=x

Prepare for math tests in your state with these Grade 3 to Grade 8 practice assessments for Common Core and state equivalents. 40 multiple choice questions and detailed answers to support test prep, created by US math experts covering a range of topics!

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DOWNLOAD FREENow, letβs look at the proof of this to understand why it is always true.

Using congruent triangles, you can prove this is always true.

First, draw two radii, AC and BC, to make two right triangles, triangle ADC and triangle BDC.

Since radii within a circle are congruent in length, you know that the hypotenuses of both right triangles (Triangle ADC and Triangle BDC ) are congruent.

Since both triangles share side CD, through the reflexive property, CD is congruent to itself.

At this point, you can conclude that right triangle ADC is congruent to the right triangle BDC through the right triangle congruence postulate, HL (Hypotenuse-Leg).

Since the two triangles are congruent, all of the corresponding parts of the triangle are congruent. Therefore, AD is congruent to BD, which will always be true.

Note: The radius is also the perpendicular bisector of the chord AB and the angle bisector of central angle BCA.

How does this apply to geometry?

**High School Geometry – Circles (HSG-C.A.2)**Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.

In order to find missing angles or the length of a chord

**Locate the key parts of the circle for an appropriate circle theorem.****Use other angle or segment facts to write an equation.****Solve for the missing side or angle.**

In circle D, if AB=15 units, find the length of AC.

**Locate the key parts of the circle for an appropriate circle theorem.**

- D is the center of the circle.
- CE is the diameter of the circle.
- CD is the radius of the circle.
- AB is the chord of the circle.
- CE is perpendicular to AB.

2**Use other angle or segment facts to write an equation.**

Using the theorem, you know that AC=BC. Using the segment addition postulate, you know that AC+BC=AB.

3**Solve for the missing side or angle.**

Given that AB=15, use substitution.

\begin{aligned} AC+BC&=15 \\\\ AC&=BC \end{aligned} \begin{aligned}BC+BC&=15 \\\\ 2BC&=15 \\\\ BC&=7.5 \end{aligned}In circle F, \, NP=24 units, and EF=FM. Find the length of DE.

**Locate the key parts of the circle for an appropriate circle theorem.**

- F is the center of a circle.
- FO and FL are radii of the circle.
- EF=FM so chord DG and chord NP are the same distance from the center of the circle.
- Radius FO is perpendicular to chord DG and radius FL is perpendicular to chord NP so the radii also bisect each of the chords.
- MP=MN and DE=EG

**Use other angle or segment facts to write an equation.**

DG=NP

NP=24 so DG=24

Since the radius bisects DG, you know that EG=DE.

GE=\cfrac{1}{2} \, (GD)

**Solve for the missing side or angle.**

\begin{aligned}EG&=\cfrac{1}{2} \, (DG) \\\\ EG&=\cfrac{1}{2} \, (24) \\\\ EG&=12\text{ units} \end{aligned}

In circle J, \, LM=121 units and KM=3x. Find the value of x.

**Locate the key parts of the circle for an appropriate circle theorem.**

- J is the center of the circle.
- LM is a chord.
- NO is the diameter.
- NO is perpendicular to chord LM and also bisects LM.

**Use other angle or segment facts to write an equation.**

Since the diameter, NO bisects chord LM, you know that LK=LM.

Using the segment addition postulate, LK+KM=LM

Substitute into the equation to get:

3x+3x=121

**Solve for the missing side or angle.**

\begin{aligned}3x+3x&=121 \\\\
6x&=121 \\\\
x&=20\cfrac{1}{6} \end{aligned}

In circle L, \, RT=18 units and PQ=12 units. Find the length of segment LS to the nearest tenth.

**Locate the key parts of the circle for an appropriate circle theorem.**

- L is the center of the circle.
- RT is the diameter of the circle.
- PQ is a chord.
- RT is perpendicular to PQ so it also bisects PQ.
- PS=QS .

**Use other angle or segment facts to write an equation.**

The diameter is 18 so the radius length is 9.

Chord PQ=12 so QS and PS are both 6 units.

Create the right triangle LSQ by drawing in another radius. The hypotenuse of the right triangle is the radius of the circle so is equal to 9 units.

Using the Pythagorean Theorem, the equation is:

9^2=6^2+(LS)^2

**Solve for the missing side or angle.**

\begin{aligned}& 9^2=6^2+(L S)^2 \\\\
& 81=36+(L S)^2 \\\\
& 45=(L S)^2 \\\\
& L S \approx 6.9 \end{aligned}

LS is approximately 6.9 units.

In circle F, the radius length is 14 units and chord AC is 22 units, find the value of BD to the nearest tenth.

**Locate the key parts of the circle for an appropriate circle theorem.**

F is the center of the circle.

BF is the radius and perpendicular to chord AC so it also bisects chord AC.

AD=CD which means each segment equals 11 units.

**Use other angle or segment facts to write an equation.**

Draw in another radius to form the right triangle CDF so that the Pythagorean Theorem can be used to find DF. Once you have the value of DF it can be subtracted from radius BF to find the length of BD.

11^2+(DF)^2=14^2 and BD=BF-FD.

**Solve for the missing side or angle.**

\begin{aligned}11^2+(DF)^2&=14^2 \\\\
121+(DF)^2&=196 \\\\
(DF)^2&=75 \\\\
DF&=8.7\text{ units} \end{aligned}

Substitute values into BD=BF-FD\text{:}

BD=14-8.7=13.96\approx{14} units

BD is approximately 14 units.

In circle O, \, AB=9 units. Find the length of BD to the nearest tenth.

**Locate the key parts of the circle for an appropriate circle theorem.**

- O is the center of the circle.
- DF is the diameter of the circle.
- DF is perpendicular to chord AB so it also bisects chord AB.
- BC=AC
- Triangle BCD is a right triangle where angle D of the triangle is 35^{\circ}.

**Use other angle or segment facts to write an equation.**

Using right triangle trigonometry, you can solve to find the length of BD.

AB=9 and BC=AC, so BC=AC=9\div{2}=4.5

In right triangle BCD, the side opposite the 35^{\circ} is known to be 4.5 units and BD is the hypotenuse. The adjacent side CD is unknown but not needed to solve this problem.

The trigonometric function that relates the angle to the opposite side and the hypotenuse is sine (\sin\theta=\cfrac{O}{H}), so use the sine ratio to solve for BD.

\sin{35}=\cfrac{4.5}{BD}

**Solve for the missing side or angle.**

Multiply both sides of the equation by BD\text{:}

BD\times\sin{35}=4.5

Divide both sides by 35 to make BD the subject of the equation:

BD=\cfrac{4.5}{\sin{35}}

Use a calculator in degree mode to find the value.

BD=\cfrac{4.5}{\sin{35}}\approx{7.8} units.

BD is approximately 7.8 units.

- Provide opportunity for students to discover the theorems and make conjectures using compasses and protractors.

- Use digital platforms so students can investigate strategies to problem solve using the circle theorems.

**Thinking that any chord that is perpendicular to another chord bisects it**

When the diameter is perpendicular to a chord it will always bisect that chord.

But not all chords that are perpendicular bisect each other.

In circle C, \, DE is the diameter which is perpendicular to chord FG. \, FG is also being bisected by DE. In circle B, chords HK and IJ are perpendicular, neither one is being bisected.

**Not using triangles to find missing lengths**

For example, in the figure below depending on the given information, it might be helpful to sketch an isosceles triangle because segments connecting CG and CH are radii of the circle and equal in length.

**Using Pythagorean theorem (Pythagorasβ theorem) incorrectly**

The missing side is calculated by incorrectly adding the square of the hypotenuse and a shorter side,**or**subtracting the square of the shorter sides.

For example, in the figure below finding x by writing the equation, 8^2+5^2=x^2 instead of x^2+5^2=8^2.

**Using the incorrect trigonometric function**

The incorrect trigonometric function is used and so the side or angle being calculated is incorrect. This also includes the inverse trigonometric functions.

For example, using the cosine function to find segment BC instead of the sine function.

1. In circle F, radius FG is perpendicular to chord AC. The length of the chord AC is 36 units and line segment AB is 3x. Find the value of x.

8

6

12

4

Since the radius is perpendicular to the chord it bisects the chord, meaning that AB=BC (so they both equal 3x ).

To find the value of x, you can apply the segment addition postulate,

\begin{aligned}3x+3x&=36 \\\\ 6x&=36 \\\\ x&=6 \end{aligned}

2. In circle G, \, GI is perpendicular to AD. If AB=3x-1 and BD=2x+7, find the length of AD.

23 units

8 units

24 units

46 units

GI is perpendicular to AD which means it also bisects AD, meaning that AB=BD.

The equation to solve for x is:

3x-1=2x+7

Subtract 2x from both sides of the equation

x-1=7

Add 1 to both sides of the equation to isolate x.

x=8

Substitute the value of x into the expression to find the value of AD.

AD=BD=3(8)-1=24-1=23

2(AB)=AD so AD=2\times{23}=46 units

3. In circle C, the radius is 10 units and AD is 12 units. Find the length of CF.

6 units

8 units

10 units

4 units

Draw in another radius to create the right triangle AFC.

Then using the Pythagorean Theorem, solve for CF.

The radius is the hypotenuse of the triangle which is 10 units.

AF is half the measure of AD so it is 6 units (\cfrac{1}{2}\times{12}=6)

\begin{aligned}(CF)^2+6^2&=10^2 \\\\ (CF)^2+36&=100 \\\\ (CF)^2&=64 \\\\ CF&=8 \end{aligned}

CF=8 units

4. In circle G, \, RT is perpendicular to AB. If RT=22 and AB=16, find HT. Round the answer to the nearest tenth.

3.5 units

7.5 units

8 units

11 units

The diameter, RT, is perpendicular to chord AB which means that it also bisects AB making AH=BH.

If AB=16, then AH=BH=\cfrac{1}{2}\times{16}=8

The diameter length is 22 so the radius length AG=\cfrac{1}{2}\times{22}=11.

AG=11

Draw in a radius to make the right triangle GAH.

Using the Pythagorean theorem, you can write the equation,

\begin{aligned}(GH)^2+8^2&=11^2 \\\\ (GH)^2+64&=121 \\\\ (GH)^2&=57 \\\\ GH&\approx{7.5} \end{aligned}

GT is a radius of circle G.

The radius length is 11, from the center of the circle at point G to H is 7.5.

To find HT subtract the two lengths.

\begin{aligned}HT&=GT-GH \\\\ HT&=11-7.5 \\\\ HT&=3.5 \end{aligned}

HT=3.5 units

5. In circle O, find the length of AE, if BD is perpendicular to AC and chord CD=8\sqrt{3}.

2\sqrt{3} units

8 units

4\sqrt{3} units

4 units

The diameter, BD is perpendicular to chord AC which means it also bisects AC, so AE=CE.

Using the right triangle CED, you can solve for x which is CE.

The right triangle is a 30-60-90 triangle with the hypotenuse, CD, being equal to 8\sqrt{3}.

Using the ratios of the sides of a 30-60-90, you can solve for x.

\cfrac{\text{short side}}{\text{hypotenuse}}=\cfrac{1}{2}=\cfrac{x}{8\sqrt{3}}

\begin{aligned}\cfrac{1}{2}&=\cfrac{x}{8\sqrt{3}} \\\\ 2x&=8\sqrt{3} \\\\ x&=4\sqrt{3} \end{aligned}

Segment CE=x, so CE=4\sqrt{3}

Since AE=CE, then AE=4\sqrt{3}

6. AC is the diameter of circle O and is perpendicular to BD. If segment DE is 7\mathrm{~cm} and angle BAE=22^{\circ}, find the value of x to the nearest tenth.

18.7\mathrm{~cm}

2.6\mathrm{~cm}

7.5\mathrm{~cm}

6.5\mathrm{~cm}

Since AC is the diameter of a circle and perpendicular to chord BD, it also bisects BD, so BE=DE.

You are given that DE=7\mathrm{~cm} which means that BE=7\mathrm{~cm}.

Using the right triangle BEA, use the sine ratio to find the value of x.

\begin{aligned}\sin(22)&=\cfrac{7}{x} \\\\ x\sin(22)&=7 \\\\ x=\cfrac{7}{\sin(22)}&\approx{18.7} \\\\ AB&=18.7\mathrm{~cm} \end{aligned}

The secant line is a line that intersects a circle at two points. The two points where the secant intersects a circle are on the edge of the circle or the circumference of the circle. Chords are part of a secant line.

The arc length is the actual distance of length of an arc of a circle. It is a portion of the circumference of the circle. Not to be confused with the degree measurement of the arc.

Equal chords of a circle subtend equal arcs (minor arcs) not angles.

- Types of data
- Averages and range
- Representing data

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[FREE] Common Core Practice Tests (3rd to 8th Grade)

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