Use the Baudhāyana–Pythagoras theorem to show why chords of a circle having the same length are all at the same distance from the centre of the circle.
Step-by-step solution
To find: Show, using the Baudhāyana–Pythagoras theorem, that CE = CH (equal chords are equidistant from the centre)
Idea: The result to be shown is: chords of a circle having the same length are all at the same distance from the centre. The radius, half the chord and the distance from the centre form a right triangle, so the distance is fixed by the radius and the chord length alone.
- Let AB = FG = l, and let E, H be their midpoints. Then CE ⟂ AB and CH ⟂ FG (line from centre to midpoint of a chord), and AE = FH = l/2.1 mark
- In right ΔCEA, by the Baudhāyana–Pythagoras theorem: CA2 = CE2 + AE2, so CE2 = r2 − (l/2)2.½ mark
- In right ΔCHF: CF2 = CH2 + FH2, so CH2 = r2 − (l/2)2.½ mark
- So CE2 = CH2, and since lengths are positive, CE = CH. Equal chords are at the same distance from the centre.1 mark
Check: Radius 5 cm, two chords of 8 cm: each distance is √(25 − 16) = 3 cm ✓.
Answer to write in the exam
Let AB = FG = l; E, H midpoints; CA = CF = r.
CE ⟂ AB, CH ⟂ FG (line from centre to midpoint of a chord); AE = FH = l/2
In right ΔCEA: CE2 = CA2 − AE2 = r2 − (l/2)2 (Baudhāyana–Pythagoras theorem)
In right ΔCHF: CH2 = CF2 − FH2 = r2 − (l/2)2
⇒ CE2 = CH2
∴ CE = CH, i.e. equal chords are equidistant from the centre.
Common mistakes that cost marks
- Using the whole chord in the right triangle: CE2 = r2 − l2. The right triangle has half the chord as one leg.
- Not explaining why E is the foot of the perpendicular: the line from the centre to the midpoint of a chord is perpendicular to it.
- Concluding CE = CH from CE2 = CH2 without noting that both are positive lengths.
How this can come in the exam
In a circle of radius 6.5 cm, two chords each of length 12 cm are drawn. The distance of each from the centre is
- 2.5 cm
- 5.5 cm
- 6 cm
- 0.5 cm
Show answer
(A) 2.5 cm
Distance = √(6.52 − 62) = √6.25 = 2.5 cm for both.
Try one yourself
A circle has radius 61 mm. Two chords of length 22 mm are drawn. Using the Baudhāyana–Pythagoras theorem, find the distance of each from the centre.
Show answer
Half-chord = 11 mm, so distance = √(612 − 112) = √(3721 − 121) = √3600 = 60 mm for each chord.
More questions like this
- Consider the figure. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.
- 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.
- You have two chords on a circle. One is longer than the other. Which chord is closer to the centre? Can you guess?
Activity: Draw a circle. Draw chords of various lengths. Drop a perpendicular to each chord from the centre. Record the length of the chord and its distance from the centre in a table.
What do you observe? - Let AB and DE be two chords of a circle with centre C. Suppose AB > DE. Then the distance from C to AB is less than the distance from C to DE.
- Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.