In the given figure, the radii of two concentric circles are 13 cm and 8 cm. AB is diameter of the bigger circle. BD is the tangent to the smaller circle touching it at D. Find the length AD.

Given: In the given figure, the radii of two concentric circles are 13 cm and 8 cm. AB is diameter of the bigger circle. BD is the tangent to the smaller circle touching it at D

To find: Length of AD

Construction: Join OD, AE and AD

Observations:

OD ⊥ BE [Tangent at a point on circle is perpendicular to the radius through point of contact]

∠AEB = 90° [Angle in a semicircle is a right-angle]

Now, In ΔABE and ΔOBD

∠AEB = ∠ODB [Both 90°]

⇒ OD || AE [Lines having corresponding angles are equal are parallel]

Also, O is the mid-point of AB and D is the mid-point of BE

∴ By Thales’s theorem

AE = 2OD

⇒ AE = 2(8) = 16 cm [As OD is radius of smaller circle]

In ΔOBD

OD = 8 cm

OB = 13 cm

By Pythagoras theorem [Hypotenuse² = Base² + Perpendicular²]

⇒ OB² = OD² + BD²

⇒ 13² = 8² + BD²

⇒ 169 – 64 = BD²

⇒ BD = √105 cm

Also, BD = DE [Perpendicular to a chord from center bisects the chord]

⇒ DE = √105 cm

Now, In ΔADE, again by Pythagoras theorem

AD² = DE² + AE²

⇒ AD² = (√105)² + 16²

⇒ AD² = 105 + 256

⇒ AD² = 361

⇒ AD = 19 cm