Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is 2√(r2 − d2).
Step-by-step solution
To find: Show that AB = 2√(r2 − d2)
Idea: Two facts together: the perpendicular from the centre bisects the chord, and the Baudhāyana–Pythagoras theorem in the right triangle formed by the radius, the distance and half the chord.
- Let AB be a chord of a circle with centre O and radius r, and let OM ⟂ AB with OM = d.½ mark
- The perpendicular from the centre to a chord bisects it, so AM = MB and AB = 2MB.1 mark
- ΔOMB is right-angled at M, with hypotenuse OB = r. By the Baudhāyana–Pythagoras theorem, OB2 = OM2 + MB2, so MB2 = r2 − d2 and MB = √(r2 − d2).1 mark
- Hence AB = 2MB = 2√(r2 − d2). (This needs d < r; at d = 0 it gives the diameter 2r.)½ mark
Check: r = 5, d = 3: 2√(25 − 9) = 2 × 4 = 8, and indeed a chord of 8 in a circle of radius 5 is 3 from the centre (32 + 42 = 52) ✓.
Answer to write in the exam
Let OM ⟂ AB, OM = d, OB = r.
AM = MB (perpendicular from the centre bisects the chord) ⇒ AB = 2MB
OB2 = OM2 + MB2 (Baudhāyana–Pythagoras theorem in ΔOMB)
⇒ MB2 = r2 − d2 ⇒ MB = √(r2 − d2)
∴ AB = 2√(r2 − d2)
Common mistakes that cost marks
- Writing √(r2 − d2) as r − d. A square root of a difference is not the difference of square roots.
- Forgetting the factor 2: √(r2 − d2) is only half the chord.
- Skipping the reason that M is the midpoint (the perpendicular from the centre bisects the chord).
How this can come in the exam
Using chord length = 2√(r2 − d2), a chord through the centre (d = 0) has length
- 0
- r
- 2r
- r2
Show answer
(C) 2r
2√(r2 − 0) = 2r, the diameter.
A chord is 11 cm from the centre of a circle of radius 61 cm. Find its length.
Show answer
Length = 2√(612 − 112) = 2√(3721 − 121) (1 mark) = 2√3600 = 2 × 60 = 120 cm (1 mark).Try one yourself
Use the formula to find the length of a chord 12 cm from the centre of a circle of radius 37 cm.
Show answer
2√(372 − 122) = 2√(1369 − 144) = 2√1225 = 2 × 35 = 70 cm.
More questions like this
- In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD = 2 AB? Give reasons for your answer.
- A circle with centre O is drawn, and A, B, C, D are points on the circle (see the figure). Measure the angles subtended by arc AKB and arc CLD at the centre O. If the angle at the centre is less than 180°, it is a minor arc. If the angle at the centre is greater than 180°, it is a major arc. State whether arcs AKB and CLD are minor arcs or major arcs.
- Draw a circle and a chord AB. Fix an arc AKB formed by AB and a point K between A, B on the circle. Measure the angle subtended at the centre by arc AKB. Take three points P, Q, R on the circle outside arc AKB. Measure the angles subtended by arc AKB at points P, Q, R. What do you notice?
Repeat this activity for a different arc AKB. - The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc.
- The angle subtended by a diameter at any point on the circle is 90°.