Answer :

__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 [Hypotenuse^{2} = Base^{2} + Perpendicular^{2}]

⇒ OB^{2} = OD^{2} + BD^{2}

⇒ 13^{2} = 8^{2} + BD^{2}

⇒ 169 – 64 = BD^{2}

⇒ 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

AD^{2} = DE^{2} + AE^{2}

⇒ AD^{2} = (√105)^{2} + 16^{2}

⇒ AD^{2} = 105 + 256

⇒ AD^{2} = 361

⇒ AD = 19 cm

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