AB and CD are two parallel chords of a circle such that AB = 10cm and CD = 24. The chords are on opposite sides of the centre and distance between them is17cm. Find the radius of the circle
Given AB = 10 cm, CD = 24 cm and distance between their center is 17 cm.
Let OQ = y cm and radius of circle = OB = OD = x cm
Therefore, OP = (17 - y) cm
OP ⊥ AB,
(Perpendicular from the centre of the circle to a chord bisects the chord)
OQ ⊥ CD
(Perpendicular from the centre of the circle to a chord bisects the chord)
In right ∆OPB,
OB2 = OP2 + PB2 (Pythagoras Theorem)
⇒ x2 = (17 - y)2 + (5)2 .... (1)
Similarly,
In right ∆OQD,
OD2 = OQ2 + QD2 (Pythagoras Theorem)
∴ x2 = y2 + 122 ... (2)
On equating (1) and (2), we get
(17 - y)2 + (5)2 = y2 + 122
⇒ 289 + y2 -34y + 25 = y2 + 144
⇒ -34y + 314 = 144
⇒ -34y = 144 - 314
⇒ -34y = -170
⇒ y = 5
On putting the value of y in (2), we get
x2 = 52 + 122
⇒ x2 = 25 + 144
⇒ x2 = 169
⇒ x = 13 cm
Thus, radius of the circle is 13 cm.