How to Solve: Chord Properties

How to Solve: Chord Properties

(For Students Who Want to Ace Their Exam & Teachers Who Need a Ready-to-Record Script)

Introduction

"Ever seen a bridge, a Ferris wheel, or even a pizza slice? Chord properties help engineers design them—and they’ll help YOU solve circle questions in under 60 seconds on your exam!

What You Need To Know First

Before diving into chord properties, ensure you understand: 1. Basic circle terminology (radius, diameter, circumference, center). 2. Perpendicular lines and right angles (90° angles, Pythagorean theorem). 3. Congruent triangles (SSS, SAS, ASA criteria).

If any of these are shaky, review them first—chord properties build on them!

Key Vocabulary

Formulas To Know

1. Perpendicular from Center to Chord

Formula: If a perpendicular is drawn from the center of a circle to a chord, it bisects the chord.

Variables: - Let O = center of the circle. - Let AB = chord. - Let OM = perpendicular from O to AB. - Then, AM = MB.

Memorise this? ✅ MEMORISE THIS (Not always given on exam sheets.)

2. Distance from Center to Chord

Formula: The length of a chord (AB) can be found using the distance (d) from the center (O) to the chord and the radius (r):

[ AB = 2 \sqrt{r^2 - d^2} ]

Variables: - r = radius of the circle. - d = perpendicular distance from center to chord. - AB = length of the chord.

Memorise this? ✅ MEMORISE THIS (Derived from Pythagoras—examiners expect you to know it.)

3. Congruent Chords & Equal Distances

Formula: In the same circle (or congruent circles), two chords are equal in length if and only if they are equidistant from the center.

Variables: - If AB = CD, then OM = ON (where OM and ON are perpendicular distances from center O to chords AB and CD). - Conversely, if OM = ON, then AB = CD.

Memorise this? ✅ MEMORISE THIS (Critical for proofs and problem-solving.)

4. Angle at the Center vs. Angle at the Circumference

Formula: The angle subtended by a chord at the center (∠AOB) is twice the angle subtended at any point on the circumference (∠ACB) on the same side of the chord.

[ \angle AOB = 2 \times \angle ACB ]

Variables: - O = center. - A, B = endpoints of chord. - C = any point on the circumference (not on chord AB).

Memorise this? ✅ MEMORISE THIS (Given on some exam sheets, but not all—know it anyway.)

Step-by-Step Method

How to Solve Any Chord Property Problem

Follow these steps in order for every question:

1. Draw the circle and label the center (O).

2. Even if the diagram is given, redraw it quickly to avoid confusion.

3. Identify the chord(s) and mark them clearly.

4. Label endpoints (e.g., A and B for chord AB).

5. Check if a perpendicular from the center to the chord is given or needed.

6. If yes, draw it (e.g., OM ⊥ AB).

7. If not, ask: "Can I draw one to simplify the problem?"

8. Apply the perpendicular bisector property.

9. If OM ⊥ AB, then AM = MB.

10. Use this to find missing lengths.

11. Use the distance formula if needed.

12. If you know r and d, find the chord length: ( AB = 2 \sqrt{r^2 - d^2} ).

13. If you know AB and r, find d: ( d = \sqrt{r^2 - \left( \frac{AB}{2} \right)^2} ).

14. Look for congruent triangles or angles.

15. If two chords are equal, their distances from the center are equal (and vice versa).

16. If angles are involved, use the "angle at center = 2 × angle at circumference" rule.

17. Write down what you’ve found and check if it answers the question.

18. Did you find the chord length? The distance? The angle? Verify!

Worked Example Using the Steps

Problem: In a circle with radius 13 cm, a chord is 10 cm from the center. Find the length of the chord.

Solution (Step-by-Step):

1. Draw the circle and label the center (O).

2. Sketch a circle with center O.

3. Identify the chord and mark it (AB).

4. Draw chord AB somewhere in the circle.

5. Draw the perpendicular from O to AB.

6. Let OM be the perpendicular, where M is the midpoint of AB.

7. Given: OM = 10 cm (distance from center to chord).

8. Apply the perpendicular bisector property.

9. AM = MB (since OM bisects AB).

10. Use the distance formula.

11. Given: r = 13 cm, d = OM = 10 cm.
12. Formula: ( AB = 2 \sqrt{r^2 - d^2} ).
13. Plug in values: ( AB = 2 \sqrt{13^2 - 10^2} ).
14. Calculate: ( AB = 2 \sqrt{169 - 100} = 2 \sqrt{69} ).
15. Simplify: ( AB = 2 \times 8.3066 \approx 16.61 ) cm.

16. Exact answer: ( AB = 2 \sqrt{69} ) cm.

17. Check for congruent triangles or angles.

18. Not needed here, but if the question asked for angles, we’d use the central angle rule.

19. Write the final answer.

20. The length of the chord is ( 2 \sqrt{69} ) cm (or ≈ 16.6 cm).

Worked Examples

Example 1 - Basic: Find Chord Length

Problem: A circle has a radius of 5 cm. A chord is 3 cm from the center. Find the length of the chord.

Solution: 1. Draw the circle with center O. 2. Draw chord AB and perpendicular OM = 3 cm. 3. OM bisects AB, so AM = MB. 4. Use the formula: ( AB = 2 \sqrt{r^2 - d^2} ).
- ( AB = 2 \sqrt{5^2 - 3^2} = 2 \sqrt{25 - 9} = 2 \sqrt{16} = 2 \times 4 = 8 ) cm.

What we did and why: We used the perpendicular distance formula because the problem gave us r and d. The key was recognizing that the perpendicular bisects the chord.

Example 2 - Medium: Find Distance from Center

Problem: In a circle of radius 10 cm, a chord of length 16 cm is drawn. Find the distance from the center to the chord.

Solution: 1. Draw the circle with center O. 2. Draw chord AB = 16 cm and perpendicular OM. 3. OM bisects AB, so AM = MB = 8 cm. 4. Use the formula: ( d = \sqrt{r^2 - \left( \frac{AB}{2} \right)^2} ).
- ( d = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6 ) cm.

What we did and why: We rearranged the chord length formula to solve for d. The key was halving the chord length first because the perpendicular bisects it.

Example 3 - Exam Style: Prove Chords are Equal

Problem: In the diagram, O is the center of the circle. OM and ON are perpendiculars to chords AB and CD respectively. If OM = ON, prove that AB = CD.

Solution: 1. Draw the circle with center O. 2. Draw chords AB and CD, with OM ⊥ AB and ON ⊥ CD. 3. Given: OM = ON. 4. In right triangles OMA and ONC:
- OA = OC (both are radii).
- OM = ON (given).
- Both triangles are right-angled at M and N.
- By RHS (Right angle-Hypotenuse-Side), ΔOMA ≅ ΔONC. 5. Therefore, AM = CN. 6. But OM and ON bisect AB and CD (perpendicular from center bisects chord).
- So, AB = 2 × AM and CD = 2 × CN. 7. Since AM = CN, AB = CD.

What we did and why: We used congruent triangles to prove the chords are equal. The key was recognizing that equal distances from the center imply equal chord lengths (and vice versa).

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, let’s lock this in for your exam. Here’s the cheat sheet:

1. Perpendicular from center? It always bisects the chord. Half the chord, then use Pythagoras.
2. Chord length formula: ( AB = 2 \sqrt{r^2 - d^2} ). Radius squared minus distance squared, square root, times two.
3. Equal chords? They’re the same distance from the center. If OM = ON, then AB = CD.
4. Angles? Central angle is double the angle at the circumference.
5. Diagram tricks? Don’t trust it—use the numbers!

Now go crush those circle questions. You’ve got this!