Question

the radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of bigger circle. BD is a tangent to smaller circle touching at D. find the length of AD

ANS IS 19cm

Answer 1

see diagram and solution enclosed.(Attachments)



To know AD, we need to know AC & CD.  

To find them, we need BC and BD first.

Given   OA=OB = R ,   OD= r

Let the radii of circles be R & r.   R > r.     R= 13cm   r =  8cm.

Finally:   AD = √(R²+3 r²) = √(13²+3*8²) = 19 cm

Answer 2

●Given two concentric circles with center O

●AB is the Diameter of bigger circle BD is a tangent to smaller circle.

=> AO = OB = 13 cm

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Now,

●Join AE and OD

●Let BD interesect bigger circle at E

●In bigger circle AB is the Diameter,.

=> angle(AEB) = 90
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●In smaller circle O is the center and BD is the tangent at point D

OD is tangent at BD

=> angle(ODB) = 90

=> angle(ODB) = angle(AEB) = 90
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» AE || CD

=> Δ AEB ~ Δ ODB
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$= > \frac{AB}{OB} = \frac{AE}{OD} \\ \\ = > \frac{2OB}{OB} = \frac{AE}{8} \\ \\ = > AE = 16 \: cm$
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$= > OD = \frac{1}{2} AE = \frac{16}{2} \\ \\ = > OD = 8 \: cm$
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Now,

$= > DE = DB = \sqrt{ {OB}^{2} - {OD}^{2} } \\ \\ = > DB = \sqrt{ {13}^{2} - {8}^{2} } \\ \\ = > DB = \sqrt{169 - 64} = \sqrt{105}$
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Now,

●In Δ AED

=> angle(AED) = 90

=> AD² = AE² + DE²

$= > {AD}^{2} = {16}^{2} + { \sqrt{105} }^{2} \\ \\ = > {AD}^{2} = 256 + 105 = 361 \\ \\ = > AD = \sqrt{361} \\ \\ = > AD = 19 \: \: cm$
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