Shortcut Trick
The half-lengths of the two chords are 24 ÷ 2 = 12 cm and 18 ÷ 2 = 9 cm.
We observe that the distance between the parallel chords is exactly the sum of their half-lengths (12 + 9 = 21 cm).
This implies the perpendicular distances from the centre to the chords are 9 cm and 12 cm.
Using the Pythagorean theorem, the radius r = √(122 + 92).
r = √(144 + 81) = √225 = 15 cm.
∴ The correct answer is 15 cm.
Alternate Method
O
A
B
C
D
M
N
12 cm
9 cm
r
r
x
21 − x
Given:
⇒ Length of chord AB = 24 cm
⇒ Length of chord CD = 18 cm
⇒ Distance between the chords = 21 cm
Formula and Concept Used:
⇒ The perpendicular drawn from the centre of a circle to a chord bisects the chord.
⇒ Pythagoras Theorem: Hypotenuse2 = Base2 + Perpendicular2
Calculation:
⇒ Let the radius of the circle be r.
⇒ Let O be the centre, and OM and ON be the perpendiculars drawn from O to the chords AB and CD respectively.
⇒ AM = AB ÷ 2 = 24 ÷ 2 = 12 cm
⇒ CN = CD ÷ 2 = 18 ÷ 2 = 9 cm
⇒ Let OM = x, then ON = 21 − x (since the total distance is 21 cm).
⇒ In right-angled ΔOMA: r2 = x2 + 122 ... (Equation 1)
⇒ In right-angled ΔONC: r2 = (21 − x)2 + 92 ... (Equation 2)
⇒ Equating Equation 1 and Equation 2:
⇒ x2 + 144 = (21 − x)2 + 81
⇒ x2 + 144 = 441 + x2 − 42x + 81
⇒ 144 = 522 − 42x
⇒ 42x = 522 − 144 = 378
⇒ x = 378 ÷ 42 = 9 cm
⇒ Substituting x = 9 in Equation 1:
⇒ r2 = 92 + 122
⇒ r2 = 81 + 144 = 225
⇒ r = √225 = 15 cm
∴ The correct answer is 15 cm.
Additional Information
Perpendicular Bisector Theorem
The perpendicular drawn from the centre of a circle to a chord always bisects the chord into two equal halves.
Distance Between Parallel Chords
If two parallel chords of lengths 2a and 2b are on the opposite sides of the centre, the distance between them is √(r2 − a2) + √(r2 − b2).
Pythagorean Triplets
A set of three positive integers a, b, and c that satisfy a2 + b2 = c2. The numbers 9, 12, and 15 form a common Pythagorean triplet, representing the sides of a right-angled triangle.