Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. Distance between AB and CD is 6 cm.
To Prove: Radius of the circle, OP = ?
Construction: Join OP and OQ, OB and OD.
Proof: Chord AB || Chord CD.
AB = 5 cm, and CD =11 cm. OP⊥AB
∴ BP = AP = \(\frac{5}{2}\) = 2.5 cm.
OQ⊥CD
∴ CQ = QD = \(\frac{11}{2}\) = 5.5 cm.
PQ = 6 cm. (Data)
Let OQ = 2 cm then, OP = (6 – x) cm.
In ∆BPO, ∠P = 90°
As per Pythagoras theorem,
OB2 = BP2 + PO2 = (2.5)2 + (6 – x)2
= 6.25 + 36 – 12x + x2
OB2 = x2– 12x + 42.25 …………….. (i)
In ∆OQD, ∠Q= 90°
∴ OD2 = OQ2 + QD2 = (x)2 + (5.5)2
OD2 = x2 + 30.25 ……………….. (ii)
OB = OD (∵ radii of same circle)
From (i) and (ii).
x2 – 12x + 42.25
= x2 + 30.25 -12x
= 30.25 – 42.25 -12x
= -12
12x = 12
∴ x = \(\frac{12}{12}\)
∴ x = 1 cm.
From (ii), OD2 = x2 + 30.25
= (1)2 + 30.25
= 1 + 30.25
∴ OD2 = 31.25
OD = √(31.25)
∴ OD = 5.59 cm.
∴ Radius of circle OP = OD = 5.59 cm.