Q. 144.3( 8 Votes )

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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


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]


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

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