Find the value of AB if radius of circle is 10 cm and PQ=12 cm and RS=16 cm?             

                            

                 

 

 

Since radius of circle = 10cm, OS = OQ = 10cm
In ΔBOQ, ∠OBQ = 90°
and BQ = PQ/2 = 6cm  (perpendicular from centre to a chord bisects the chord)
By Pythagoras theorem, OQ² = OB² + BQ²
⇒ 10² = 6² + OB²
⇒ OB² = 100 – 36 = 64
⇒ OB = 8cm
Similarly, In ΔAOS, ∠OAS = 90°
and AS = RS/2 = 8cm  (perpendicular from centre to a chord bisects the chord)
By Pythagoras theorem, OS² = OA² + AS²
⇒ 10² = 8² + OA²
⇒ OA² = 100 – 64 = 36
⇒ OA = 6cm
Now, AB = OB – OA
⇒ AB = 8 – 6 = 2cm.