Let O be the centre of the circle and AB and CD are two parallel chords on the same side of the radius. OP is perpendicular to AB and OQ is perpendicular to CD. If AB = 10 cm, CD = 24 cm and PQ = 7 cm, then the diameter (in cm) of the circle is equal to:
Show answer & explanation
Correct answer: A
Explanation
The correct answer is Option 1 i.e. 26.
Given that, AB = 10 cm, AP = 5 cm
CD = 24 cm, QC = 12 cm
PQ = 7 cm = OP - OQ ...(i)
Radius of the circle OC = OA = r cm
Now, in triangle OQC, using Pythagoras' theorem,
⇒ OC2 = OQ2 + QC2
⇒ r2 = OQ2 + 144 ...(ii)
Also in triangle OPA, using Pythagoras' theorem,
⇒ OA2 = OP2 + AP2
⇒ r2 = OP2 + 25 ...(iii)
Equating equations (ii) and (iii),
⇒ OQ2 + 144 = OP2 + 25
⇒ OP2 - OQ2 = 119
⇒ (OP + OQ)(OP - OQ) = 119
⇒ OP + OQ = 17 ...(iv)
Using equations (i) and (iv),
OP = 12 and OQ = 5
Now putting this value in equation (iii),
⇒ r2 = 122 + 25
⇒ r2 = 144 + 25
⇒ r = 13 cm
Diameter = 2r = 26 cm