If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF. Solve this using the Baudhāyana–Pythagoras theorem.
Step-by-step solution
To find: Show that AB = GF using the Baudhāyana–Pythagoras theorem
Idea: The problem: CE is perpendicular to chord AB, CH is perpendicular to chord GF, and CE = CH; show AB = GF. This time, instead of congruent triangles, work out each half-chord from the right triangle radius–distance–half-chord.
- In right ΔCEA (∠E = 90°): CA2 = CE2 + AE2, so AE2 = r2 − CE2.1 mark
- In right ΔCHF (∠H = 90°): FH2 = r2 − CH2.½ mark
- CE = CH, so AE2 = FH2, giving AE = FH.½ mark
- The perpendicular from the centre bisects a chord: AB = 2AE and GF = 2FH. So AB = GF.1 mark
Check: Radius 13 cm with both chords 12 cm from the centre: each half-chord is √(169 − 144) = 5 cm, so both chords are 10 cm ✓.
Answer to write in the exam
Let the radius be r: CA = CF = r.
AE2 = CA2 − CE2 = r2 − CE2 (Baudhāyana–Pythagoras theorem in ΔCEA)
FH2 = CF2 − CH2 = r2 − CH2 (in ΔCHF)
CE = CH (given) ⇒ AE2 = FH2 ⇒ AE = FH
AB = 2AE, GF = 2FH (perpendicular from the centre bisects the chord)
∴ AB = GF
Common mistakes that cost marks
- Writing AE2 = r2 + CE2. The radius is the hypotenuse, so subtract.
- Forgetting that AB is twice AE; the theorem gives only the half-chord.
- Assuming AB = GF to find the half-chords; the equal distances are what is given.
How this can come in the exam
Chord PQ of a circle of radius 41 cm is 9 cm from the centre. Chord RS is also 9 cm from the centre. Find RS.
Show answer
Half of RS = √(412 − 92) = √(1681 − 81) = √1600 = 40 cm (1 mark). RS = 80 cm, the same as PQ (1 mark).Try one yourself
In a circle of radius 89 cm, two chords are each 39 cm from the centre. Find their lengths.
Show answer
Half-chord = √(892 − 392) = √(7921 − 1521) = √6400 = 80 cm, so each chord is 160 cm.
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