Question

A chord of a length 16.8 cm is at a distance of 11.2 cm from the centre of a circle . Find the length of the chord of the same circle which is at a distance of 8.4 cm from the centre.

Answer 1

AF = FB = 8.4 cm 

And DE = EC ----(1)
(Perpendicular from centre to a chord bisects the chord) 

In right Δ ODA, 

By Pythagoras theorem, OA² = OF² + AF² 

= (11.2)² + (8.4)²

= 125.44 + 70.56

OA² = 196

OA = 14 cm

OA = OD = 14cm    (radii of same circle) 

Similarly, In Δ  DEO 

OD² = OE² + DE² 

DE² = 14² + 8.4² 

= 196 - 70.56

DE² = 125.44

DE = 11.2 cm

∴  length of chord DC = 2DE = 2(11.2) = 22.4 cm