In the figure, O is the centre of the circle of radius 10 cm. If PQ = 12 cm, RS = 16 cm, then find the value of AB.

Given: Radius of circle = 10 cm

PQ = 12 cm

RS = 16 cm

Let's join point O with S and Q.

∴ OS = OQ = 10 cm   [Radii of a circle]

Since perpendicular from the centre of the circle to a chord bisects the chord.

$\therefore BQ=\frac{PQ}{2}=\frac{12}{2}=6 cm$

And, $AS=\frac{RS}{2}=\frac{16}{2}=8 cm$

In △OBQ, ∠OBQ = 90^o

∴ OQ² = OB² + BQ²     [By Pythagoras theorem]

⇒ OB² = OQ² - BQ²

⇒ OB² = (10)² - (6)²

⇒ OB² = 100 - 36 = 64

⇒ OB = 8 cm

Similarly, In △OAS, ∠OAS = 90^o

∴ OS² = OA² + AS²     [By Pythagoras theorem]

⇒ OA² = OS² - AS²

⇒ OA² = (10)² - (8)²

⇒ OA² = 100 - 64 = 36

⇒ OA = 6 cm

Now, AB = OB - OA

⇒ AB = 8 - 6

⇒ AB = 2 cm

Hence, the value of AB is 2 cm.