Lessons
Grade 10
Grade 11
Chords of a Circle
Grade
10
Term
3
This lesson is about the powerful relationship between the centre of a circle, its chords, and right angles. A chord is any straight line that connects two points on the circumference. The two theorems in this lesson are your primary tools for solving problems involving chords.
1. The Two Key Theorems (Core Concepts)
These two theorems are opposites (converses) of each other. Knowing which one to use depends on what information is given in the problem.
Given Information | Resulting Property (The Theorem) | What it Means |
You are told a line from the centre goes to the midpoint of a chord. | The line is perpendicular to the chord. | ∠OXA = 90° |
You are told a line from the centre is perpendicular to a chord. | The line bisects the chord (cuts it in half). | AX = XB |
In short:
Midpoint → 90° Angle
90° Angle → Midpoint
2. The Main Application: Creating Right-Angled Triangles
The number one strategy for almost every calculation problem in this topic is to create a right-angled triangle and then use Pythagoras' Theorem.
How to create the triangle:
Draw the radius from the centre to one end of the chord. This is your hypotenuse.
Draw the perpendicular line from the centre to the chord. This is one side.
The third side is half the length of the chord.
Problem-Solving Game Plan for Calculations:
Draw a Sketch: Draw a clear diagram of the circle, its centre O, and the chord AB.
Identify the Given Info: Read the question. Are you given the chord length? The radius? The distance from the centre to the chord (OX)?
Apply the Correct Theorem:
If given the distance (OX), you know it's perpendicular, so you can halve the chord length to find AX.
If given the midpoint, you know you have a 90° angle.
Form the Right-Angled Triangle: Focus on the triangle ΔOXA.
Apply Pythagoras' Theorem: (Radius)² = (Distance)² + (Half Chord)² or OA² = OX² + AX².
Solve for the unknown length.
Worked Example: Question: A chord of length 16 cm is 6 cm away from the centre of a circle. Find the radius.
Sketch: Draw circle with centre O, chord AB=16, and line OX=6 perpendicular to AB.
Theorem: The perpendicular from the centre bisects the chord.
Calculate Half-Chord: AX = 16 / 2 = 8 cm.
Pythagoras: In ΔOXA, OA is the radius.
OA² = OX² + AX²
Radius² = 6² + 8² = 36 + 64 = 100
Radius = √100 = 10 cm.
3. Strategy for Proofs (Riders)
Proofs in this topic almost always involve proving two triangles are congruent.
Join the Dots: Your first step is often to draw in the radii from the centre to the ends of the chords (e.g., join OA, OB, OC, etc.). This is crucial.
Identify the Right-Angled Triangles: Use the chord theorems to establish that you have right-angled triangles.
Prove Congruence: The most common method is RHS (Right Angle, Hypotenuse, Side).
R (Right Angle): Comes from Theorem 1.
H (Hypotenuse): This is always a radius, and all radii are equal.
S (Side): This is often half a chord length or a given length.
State Conclusion: Use the fact that "corresponding sides/angles of congruent triangles are equal" to complete your proof.
A Very Common Rider: Prove that equal chords are equidistant from the centre. (If AB = CD, prove OX = OY). You do this by proving ΔOXB ≅ ΔOYD using RHS.
4. Important Points to Remember
Keywords are Clues:
"Midpoint" → Your brain should think: 90° angle.
"Perpendicular" or "shortest distance" → Your brain should think: Chord is bisected.
State Full Reasons: In a proof, do not just write "Chord Theorem". You must write the full statement: (Perpendicular from the centre to a chord bisects the chord).
වියාචනය (Disclaimer)
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ජාතික විභාග සඳහා අන්තර්ගතයේ නිල බලය ලත් මූලාශ්රය වනුයේ රජය විසින් නිකුත් කරනු ලබන මෙම ප්රකාශනයි.
