Unit 8.2  Properties of chords in circles
what we need to know
 Radius vs. diameter in circles.
 Definition of a chord (don't worry! I'll be reviewing this shortly!).
 Definition of a bisector line.
 How to identify the hypotenuse and the legs of a right triangle.
 The formulas to use when using the Pythagorean Theorem.
 ISOSCELES TRIANGLES > 2 EQUAL SIDES > 2 EQUAL ANGLES.
 EQUILATERAL TRIANGLES > 3 EQUAL SIDES > 3 EQUAL ANGLES.
 The SUM of all angles inside a triangle IS ALWAYS 180 degrees.
 Definition of a chord (don't worry! I'll be reviewing this shortly!).
 Definition of a bisector line.
 How to identify the hypotenuse and the legs of a right triangle.
 The formulas to use when using the Pythagorean Theorem.
 ISOSCELES TRIANGLES > 2 EQUAL SIDES > 2 EQUAL ANGLES.
 EQUILATERAL TRIANGLES > 3 EQUAL SIDES > 3 EQUAL ANGLES.
 The SUM of all angles inside a triangle IS ALWAYS 180 degrees.
what we talked about in class  extra information
Take a look at the following definitions. Radius and diameter should be familiar, whereas a chord is the main definition you have to be able to know for this part of the unit:
Property of chords in circles
As described above, a CHORD is a line segment that joins two points on a circle. That is, a CHORD is a line that goes from side to side that, unlike the diameter, it does not go through the center of the circle.
In this part of Unit 8 we will use the property of chords in circles to find out the lengths of either the hypotenuse or the legs of right triangles (drawn inside a circle). Take a look at the following explanation:
In this part of Unit 8 we will use the property of chords in circles to find out the lengths of either the hypotenuse or the legs of right triangles (drawn inside a circle). Take a look at the following explanation:
... And what does that mean?
1. A bisector line means that this line cuts the chord in 2 EQUAL PARTS.
2. Anytime you are dealing with a chord, you can draw a line from the center of the circle TO the center of the chord. This line is the bisector line.
3. Be careful! Because this "bisector" comes from the center but it stops at the CHORD (does not go all the way to the outside of the circle), this line is NOT A RADIUS.
4. It turns out... You can ALWAYS draw a line that goes through the center TO the middle of the chord. This line is a bisector, as previously said, which means that the CHORD IS CUT in 2 EQUAL PARTS.
5. The bisector line and the chord are, then, ALWAYS PERPENDICULAR TO EACH OTHER.
6. Because perpendicular lines are "90 degrees" to each other, the angles formed by the bisector line AND each of the halves of the chord ARE ALWAYS 90 DEGREES (RIGHT ANGLES!).
7. And since we are dealing with RIGHT ANGLES, we can use... you guessed it!.... the PYTHAGOREAN THEOREM to find the information we are asked to find.
Before going on, let's take one more look at the formulas and information involved with the Pythagorean Theorem:
2. Anytime you are dealing with a chord, you can draw a line from the center of the circle TO the center of the chord. This line is the bisector line.
3. Be careful! Because this "bisector" comes from the center but it stops at the CHORD (does not go all the way to the outside of the circle), this line is NOT A RADIUS.
4. It turns out... You can ALWAYS draw a line that goes through the center TO the middle of the chord. This line is a bisector, as previously said, which means that the CHORD IS CUT in 2 EQUAL PARTS.
5. The bisector line and the chord are, then, ALWAYS PERPENDICULAR TO EACH OTHER.
6. Because perpendicular lines are "90 degrees" to each other, the angles formed by the bisector line AND each of the halves of the chord ARE ALWAYS 90 DEGREES (RIGHT ANGLES!).
7. And since we are dealing with RIGHT ANGLES, we can use... you guessed it!.... the PYTHAGOREAN THEOREM to find the information we are asked to find.
Before going on, let's take one more look at the formulas and information involved with the Pythagorean Theorem:
example problems and strategy to use
Let's take a look at the following problems:
 Always start by identifying the chord and its bisector. The chord is AB and its bisector is OC.
 Then, DRAW A TRIANGLE. This step is of great importance.. So much so, I am going to say: WHEN IN DOUBT, DRAW A TRIANGLE.
 In the example above. we needed to draw a line from O to B to complete a RIGHT TRIANGLE. Notice that, because this line goes from the center of the circle TO the outside of the circle, THIS LINE IS A RADIUS.
 Also, because this line is ACROSS FROM THE RIGHT ANGLE, line OB is the HYPOTENUSE of the drawn triangle.
 Now that we have the length of the bisector line OC (a "leg"), and the HYPOTENUSE (the radius) OB, we can find the bottom leg of the triangle, which is line CB.
 We can use the property of chords, the one that tells us that CB = CA to find the CHORD AB. That is, once we find CB using the Pythagorean Theorem, we can multiply it by 2 to find the length of the chord AB.
 Then, DRAW A TRIANGLE. This step is of great importance.. So much so, I am going to say: WHEN IN DOUBT, DRAW A TRIANGLE.
 In the example above. we needed to draw a line from O to B to complete a RIGHT TRIANGLE. Notice that, because this line goes from the center of the circle TO the outside of the circle, THIS LINE IS A RADIUS.
 Also, because this line is ACROSS FROM THE RIGHT ANGLE, line OB is the HYPOTENUSE of the drawn triangle.
 Now that we have the length of the bisector line OC (a "leg"), and the HYPOTENUSE (the radius) OB, we can find the bottom leg of the triangle, which is line CB.
 We can use the property of chords, the one that tells us that CB = CA to find the CHORD AB. That is, once we find CB using the Pythagorean Theorem, we can multiply it by 2 to find the length of the chord AB.
 What do we need to find?: The CHORD JK.
 Draw a triangle by joining the center O with point K of the CHORD. This line is THE RADIUS (it goes from the center to the circumference of the circle).
 You WILL ALWAYS have two of the three side lengths of the triangle. In this case, we have LEG 1, which is 5 cm, and THE HYPOTENUSE, which happens to be the line you just drew, as well as the radius.
 BUT... you are given the diameter and NOT the radius. Lucky for us, the radius is always half the diameter. Thus, the radius is 9 cm.
 Now, you have all the information needed to find b, which is half of the chord.
 Once we have b, multiply it by 2 to find the chord JK..
 Draw a triangle by joining the center O with point K of the CHORD. This line is THE RADIUS (it goes from the center to the circumference of the circle).
 You WILL ALWAYS have two of the three side lengths of the triangle. In this case, we have LEG 1, which is 5 cm, and THE HYPOTENUSE, which happens to be the line you just drew, as well as the radius.
 BUT... you are given the diameter and NOT the radius. Lucky for us, the radius is always half the diameter. Thus, the radius is 9 cm.
 Now, you have all the information needed to find b, which is half of the chord.
 Once we have b, multiply it by 2 to find the chord JK..
 What do we need to find? Distance X.
 What do we have?: We have a right triangle with a hypotenuse (also the radius) of 20 cm, and leg a, which happens to be HALF the chord or 12 cm (See? Two out of three lengths are always given).
 Use the given triangle to find the LEG OX. Using the Pythagorean Theorem, we find OX, or b, to be 16 cm.
 HOWEVER, you may notice that DISTANCE X is a line equal to the RADIUS (a line that goes from the center O to the circle) MINUS OX. In other words, the radius is the sum of LEG B and DISTANCE X.
 Since the radius is 20 cm, and LEG B is 16 cm, this means that a distance of 4 MUST make up the rest of the radius. THUS, DISTANCE X is equal to 4 cm.
 WHAT DO WE KNOW?
We know that OC is the bisector, and that because this bisector is perpendicular to the chord AB, they meet at right angles. This means, therefore, that angle x MUST be a right angle (90 degrees).
 LOOK AT TRIANGLE OCA:
Add the 30 and the 90 degree angles. This is a total of 120 degrees. Since the sum all of angles inside a triangle is always 180, we can infer that the missing angle is 60 degrees. That is: 180  (90 + 30) = 180  120 = 60..
 LOOK AT TRIANGLE OCB:
The first thing to be noted is the fact that triangle OCB is exactly like triangle OCA. ANd as you already know, similar triangles have equal angles. This has to mean that angle Z = angle A = 30 degrees.
 ANOTHER WAY TO LOOK AT IT:
Look at the BIGGER TRIANGLE OAB. When you do, notice that OA, one of the legs, is the radius. And so is OB! This HAS TO MEAN that, because triangle OAB has two EQUAL SIDES, which makes it an ISOSCELES TRIANGLE, it means that angle Z AND angle at A (30 degrees) ARE EQUAL.
This means, therefore, that ANGLE Z = 30 degrees.
OK. That is A LOT of information. But don't worry...
FOLLOW THE FOLLOWING STRATEGY when dealing with CHORDS IN A CIRCLE:
1. DRAW (COMPLETE) A RIGHT TRIANGLE. This triangle must inlcude, as one of its legs of the hypotenuse, that which you need to find.
2. WRITE DOWN THE 2 LENGTHS YOU HAVE: When given, one of these lenghts IS THE RADIUS. If not given, then YOU ARE BEING ASKED TO FIND THE RADIUS.
3. IDENTIFY THE LENGTH YOU DO NOT HAVE: This is an important part of what you are being asked.
4. LOOK AT THE ACTUAL PROBLEM to identify exactly what it is that you are being asked to find.
5. TO DO ALL THIS: you assumed the chord is being bisected by a line coming from the center; you assumed that the chord is "cut" into two EQUAL PARTS; and you assumed that where the bisector intercepts the chord, 90 degree angles are found.
We know that OC is the bisector, and that because this bisector is perpendicular to the chord AB, they meet at right angles. This means, therefore, that angle x MUST be a right angle (90 degrees).
 LOOK AT TRIANGLE OCA:
Add the 30 and the 90 degree angles. This is a total of 120 degrees. Since the sum all of angles inside a triangle is always 180, we can infer that the missing angle is 60 degrees. That is: 180  (90 + 30) = 180  120 = 60..
 LOOK AT TRIANGLE OCB:
The first thing to be noted is the fact that triangle OCB is exactly like triangle OCA. ANd as you already know, similar triangles have equal angles. This has to mean that angle Z = angle A = 30 degrees.
 ANOTHER WAY TO LOOK AT IT:
Look at the BIGGER TRIANGLE OAB. When you do, notice that OA, one of the legs, is the radius. And so is OB! This HAS TO MEAN that, because triangle OAB has two EQUAL SIDES, which makes it an ISOSCELES TRIANGLE, it means that angle Z AND angle at A (30 degrees) ARE EQUAL.
This means, therefore, that ANGLE Z = 30 degrees.
OK. That is A LOT of information. But don't worry...
FOLLOW THE FOLLOWING STRATEGY when dealing with CHORDS IN A CIRCLE:
1. DRAW (COMPLETE) A RIGHT TRIANGLE. This triangle must inlcude, as one of its legs of the hypotenuse, that which you need to find.
2. WRITE DOWN THE 2 LENGTHS YOU HAVE: When given, one of these lenghts IS THE RADIUS. If not given, then YOU ARE BEING ASKED TO FIND THE RADIUS.
3. IDENTIFY THE LENGTH YOU DO NOT HAVE: This is an important part of what you are being asked.
4. LOOK AT THE ACTUAL PROBLEM to identify exactly what it is that you are being asked to find.
5. TO DO ALL THIS: you assumed the chord is being bisected by a line coming from the center; you assumed that the chord is "cut" into two EQUAL PARTS; and you assumed that where the bisector intercepts the chord, 90 degree angles are found.
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review  workbook
workbook__unit_8.2.pdf  
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