Shortcut Trick
For a circle with two parallel chords of lengths L1 and L2 at distances d1 and d2 from the center, the radius r satisfies: r2 = d12 + (L1/2)2 = d22 + (L2/2)2.
Given L1 = 10 (half = 5), d1 = 12 ⇒ r2 = 122 + 52 = 169 ⇒ r = 13 cm.
For L2 = 24 (half = 12) ⇒ 132 = d22 + 122 ⇒ 169 = d22 + 144 ⇒ d2 = 5 cm.
Since chords are on opposite sides of the center, Distance = d1 + d2 = 12 + 5 = 17 cm.
∴ The correct answer is 17 cm.
Alternate Method
Given:
Length of chord AB = 10 cm, Chord CD = 24 cm.
Distance of AB from center (OM) = 12 cm.
Formula Used:
The perpendicular from the center of a circle to a chord bisects the chord. In a right-angled triangle, Pythagoras theorem states: Hypotenuse2 = Base2 + Perpendicular2.
In circle with center O, let AB and CD be the parallel chords.
OM ⊥ AB and ON ⊥ CD. Since they are on opposite sides, the distance between them is MN = OM + ON.
⇒ AM = AB ÷ 2 = 10 ÷ 2 = 5 cm.
⇒ In ΔOAM, OA2 = OM2 + AM2
⇒ r2 = 122 + 52 = 144 + 25 = 169
⇒ r = √169 = 13 cm.
⇒ In ΔOCN, OC2 = ON2 + CN2, where CN = CD ÷ 2 = 24 ÷ 2 = 12 cm.
⇒ 132 = ON2 + 122
⇒ 169 = ON2 + 144
⇒ ON2 = 169 − 144 = 25
⇒ ON = 5 cm.
⇒ Total distance MN = OM + ON = 12 + 5 = 17 cm.
∴ The correct answer is 17 cm.
Additional Information
Distance Between Parallel Chords
If chords are on the same side of the center, distance = |d1 − d2|. If on opposite sides, distance = d1 + d2.
Perpendicular Bisector Property
The perpendicular from the center to a chord always bisects it, creating two right-angled triangles with the radius as the hypotenuse.
Pythagorean Triplets
Common triplets like (5, 12, 13) and (8, 15, 17) frequently appear in circle geometry problems involving chords and radii.