Answer:

Given :

A Circle ( O,r ) having r = 5cm , chord AB and CD of length 6cm and 8cm respectively . AB parallel to CD .

To Find : PQ .

Construction : Join AO and CO .

Step Explanation :

Solution : We know that line passing from centre bisects the chord in two parts at right angle .

So , AP = 3cm

and CQ = 4cm .

So , Applying Pythagorean theorem in ∆APO and ∆CQO respectively , we get

1) AO² = AP² + PO²

5² = 3² + PO²

PO = 4cm . --- (a)

2) CO² = OQ² + CQ²

5² = 4² + OQ²

OQ = 3cm . -----(b)

So , Adding a and b , we get

PQ = (3+4)cm = 7cm .

Solution Tip : You can shift PQ to pass through Centre of the Circle (O,r) .