Data: Chords of a circle are 6 cm. and 8 cm. are parallel. The smaller chord is at distance of 4 cm. from the centre.
To Prove: Distance between bigger chord and centre = ?
Construction : Join OA and OC.
Proof: AB || CD, AB = 6 cm, CD = 8 cm.
OP⊥CD, OQ⊥CD.
In ∆OPA, ∠P = 90°
∴ OA2 = OP2 + PA2 (According to Pythagoras theorem)
= (4)2+ (3)2 = 16 + 9
OA2 = 25
∴ OA = 5 cm. OA = OC = 5 cm.
(radii of the same circle.)
Now, in ∆OQC,
OC2 = OQ2 + QC2 (5)2 = x2 + (4)2
25 = x2 + 16
x2 = 25 – 16 = 9
∴ x = √9
∴ x = 3 cm.
∴ Bigger chord is at a distance of 3 cm. from the centre.