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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.
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Chord Theorems (perpendicular from Centre)
What is it?
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.
In simple words
- 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
Important terms
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.
What you should learn
- 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.
Examples
- Radius to midpoint
- Right triangle OMA
- OA is hyp
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.
Recommended for Class 10
Full school-level explanation
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.
Midpoint and right angle travel together:
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.
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.
Remember
- M midpoint => OM perp AB
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.
Remember
- foot is midpoint
Examples
- AM=MB
- Not 'doubles the chord'
- M is on AB
Remember
- bisect <=> perpendicular from O
- 12 cm
- radius is hyp
- M midpoint => OM perp AB
- foot is midpoint
- bisect <=> perpendicular from O
- bisect <=> perpendicular from O
- mathematical reasoning
How the ideas connect
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
Deeper understanding
Go beyond memorising Chord Theorems (perpendicular from Centre). Connect the definition to reasons, relationships, exceptions, and exam wording.
How the ideas connect
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
Common mistakes
Exam focus
Core Idea
BISECT -> PERP
PERP -> BISECT
Worked Method
Exam focus
Quick revision
- bisect <=> perpendicular from O
- 12 cm
- radius is hyp
- M midpoint => OM perp AB
- CHORD THEOREMS (PERPENDICULAR FROM CENTRE): A line from the centre that bisects a chord is perpendicular to the chord. Conversely, the perpendic
- BISECT -> PERP: If OM bisects chord AB (M midpoint) then OM perp AB. Useful when you know the midpoint and need a ri
OA is the hypotenuse in triangle OMA, not the chord. The radius goes to A, not along the chord.
For long answers on Chord Theorems (perpendicular from Centre): define clearly → explain the main parts → give one example → state one exam caution.
Exam Focus
Exam Focus
Important points
- bisect <=> perpendicular from O
- 12 cm
- radius is hyp
- M midpoint => OM perp AB
Important formulas
- bisect <=> perpendicular from O
- 12 cm
- radius is hyp
Common exam traps
- OA is the hypotenuse in triangle OMA, not the chord. The radius goes to A, not along the chord.
- Not 'doubles the chord'
Long question structure
Definition → key parts → example → exam caution
Common Mistakes
Common Mistakes
- Mistake
OA is the hypotenuse in triangle OMA, not the chord. The radius goes to A, not along the chord.
- Mistake
Not 'doubles the chord'
- Mistake
Do not treat BISECT -> PERP and PERP -> BISECT as the same step; each has a different purpose.
- Mistake
OA is the hypotenuse in triangle OMA, not the chord. The radius goes to A, not along the chord.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
bisect <=> perpendicular from O
12 cm
radius is hyp
M midpoint => OM perp AB
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:
bisect <=> perpendicular from O
- 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:
- Formula
- bisect <=> perpendicular from O
- Example
- Radius to midpoint
- Critical warning
- OA is the hypotenuse in triangle OMA, not the chord. The radius goes to A, not along the chord.
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Exam Practice
Define
Q1. Define Chord Theorems (perpendicular from Centre).
Explain
Q2. Explain: BISECT -> PERP.
Example
Q3. Give one clear example of Chord Theorems (perpendicular from Centre) and say why it fits.
Mistake
Q4. State one common mistake students make about Chord Theorems (perpendicular from Centre).
Exam note
Q5. Write a short exam note on Chord Theorems (perpendicular from Centre): definition, two points, one example.
Sources & References
School-standard facts for Matric and Intermediate. Original explanations — not copied from textbooks.