1. In the given figure, O is the centre of the circle with radius 5 cm. OP ⊥ AB, OQ ⊥ CD, AB || CD, AB = 6 cm and CD = 8 cm. Determine PQ.


2. PQ and RS are two parallel chords of a circle on the same side of centre. If the radius of the circle is 10 cm and PQ = 16 cm, Rs = 12 cm then find the distance between the chords.

 

Hello Julia, OP will be a perpendicular bisector. So join OA. OA = 5 cm. AP = 3 cm. 
In right triangle, OP^2 = 5^2 - 3^2 = 4^2 ==> OP = 4 cm
Same way in right triangle OCQ, OQ^2 = OC^2 - CQ^2 = 5^2 - 4^2 = 3^2
==> OQ = 3 cm
Hence PQ = PO + OQ = 4 + 3 = 7 cm 
For the second one, OL = ​√(OP^2 - PL^2) = √10^2 - 8^2 = √6^2 = 6 cm
Same way OM = √(OR^2 - RM^2) = √10^2 - 6^2 = 8^2
 ==> OM = 8 cm
Distance between chords = LM = OM - OL = 8 - 6 = 2 cm