MathematicsChapter 9: Chord and Arcs of a Circle
Chord Theorems (perpendicular From Centre)
A line from the centre that bisects a chord is perpendicular to the chord.
Learning objectives
- Explain chord theorems (perpendicular from centre) using the correct definitions, conditions, and notation.
- Apply bisect <=> perpendicular from O correctly in routine and exam-style questions.
- Show each important step, simplify the result, and check that it satisfies the original conditions.
Definition
A line from the centre that bisects a chord is perpendicular to the chord. Conversely, the perpendicular from the centre to a chord bisects the chord. Together: the line from O to the midpoint of a chord is the perp from O to the chord.
Mathematical form
AM = 5 cm
OM = 12 cm
AM = 5 cm OA2
Where / symbols
- CHORD THEOREMS (PERPENDICULAR FROM CENTRE)
- A line from the centre that bisects a chord is perpendicular to the chord. Conversely, the perpendicular from the centre to a chord bisects the chord. Together: the line from O to the midpoint of a chord is the perp from O to the chord.
- BISECT -> PERP
- If OM bisects chord AB (M midpoint) then OM perp AB. Useful when you know the midpoint and need a right triangle OMA.
- PERP -> BISECT
- Drop a perp from O to chord AB, foot M. Then AM=MB. Equal distances on the chord. Combine with Pythagoras for length problems.
Relationship
- bisect <=> perpendicular from O
Overview
Midpoint and right angle travel together:
Bisect -> Perp
If OM bisects chord AB (M midpoint) then OM perp AB. Useful when you know the midpoint and need a right triangle OMA.
Examples Radius to midpoint · Right triangle OMA · OA is hyp
Perp -> Bisect
Drop a perp from O to chord AB, foot M. Then AM=MB. Equal distances on the chord. Combine with Pythagoras for length problems.
Examples AM=MB · Not 'doubles the chord' · M is on AB
WHAT — CHORD THEOREMS (PERPENDICULAR FROM CENTRE)
A line from the centre that bisects a chord is perpendicular to the chord. Conversely, the perpendicular from the centre to a chord bisects the chord. Together: the line from O to the midpoint of a chord is the perp from O to the chord. The symbols and conditions are part of the mathematics, so include them when stating the rule.
WHY — Why this topic matters
Chord Theorems (Perpendicular From Centre) matters because chord and arc theorems convert circle diagrams into equal lengths, equal angles, and right triangles. They are the foundation for later tangent proofs and constructions.
HOW — How to solve and explain it
Start by identifying what is given and what must be found. Work through BISECT -> PERP, then PERP -> BISECT. Worked: radius 13 cm, chord 10 cm, distance from centre to chord. perp from O meets chord at M, so M is midpoint AM = 5 cm OA^2 = OM^2 + AM^2 13^2 = OM^2 + 5^2 169 = OM^2 + 25 OM^2 = 144 OM = 12 cm Finish by checking signs, restrictions, units, and whether the answer matches the question.
Worked example
- radius 13 cm, chord 10 cm, distance from centre to chord.
- perp from O meets chord at M, so M is midpoint
- AM = 5 cm
- OA2 = OM2 + AM2
- 132 = OM2 + 52
- 169 = OM2 + 25
- OM2 = 144
- OM = 12 cm
In simple words
OA is the hypotenuse in triangle OMA, not the chord. The radius goes to A, not along the chord.
Key points
- bisect <=> perpendicular from O
- M midpoint => OM perp AB
- foot is midpoint
Exam preparation
Exam Focus
Quick revision
Revision Zone
Review the key facts, formulas and checks before you move on.Chord Theorems (perpendicular From Centre)
Sticky Notes · Quick Revision
Core Idea
- A line from the centre that bisects a chord is perpendicular to the chord. Conversely, the perpendicular from the centre to a chord bisects the chord. Together: the line from O to the midpoint of a chord is the perp from O to the chord.
BISECT -> PERP
- If OM bisects chord AB (M midpoint) then OM perp AB. Useful when you know the midpoint and need a right triangle OMA.
PERP -> BISECT
- Drop a perp from O to chord AB, foot M. Then AM=MB. Equal distances on the chord. Combine with Pythagoras for length problems.
Worked Method
- Worked: radius 13 cm, chord 10 cm, distance from centre to chord.
- perp from O meets chord at M, so M is midpoint
- AM = 5 cm
- OA2 = OM2 + AM2
- 132 = OM2 + 52
- 169 = OM2 + 25
- OM2 = 144
- OM = 12 cm
Common Misconception
- OA is the hypotenuse in triangle OMA, not the chord. The radius goes to A, not along the chord.
FBISE Exam Check
- OA is the hypotenuse in triangle OMA, not the chord. The radius goes to A, not along the chord.
60-second revision
- A line from the centre that bisects a chord is perpendicular to the chord.
- BISECT -> PERP: If OM bisects chord AB (M midpoint) then OM perp AB.
- PERP -> BISECT: Drop a perp from O to chord AB, foot M.
- Worked-method reminder: OM = 12 cm
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