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


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


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