Question

In the given figure, quad. ABCD is circumscribed, touching the circle at P, Q, R and S such that ∠DAB = 90°, If CS = 27 cm and CB = 38 cm and radius of the circle is 10 cm, then AB = ?


(a) 28 cm
(b) 21 cm
(c) 19 cm
(d) 17 cm

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)21\mathrm{cm} \\ \mathrm{Given},CS=27\mathrm{cm},CB=38\mathrm{cm}\mathrm{and}\angle DAB={90}^0. \\ \text{T}\mathrm{he}\mathrm{}\mathrm{length}s\mathrm{}\mathrm{of}\mathrm{}\mathrm{tangents}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{from}\mathrm{an}\mathrm{external}\mathrm{}\mathrm{point}\mathrm{}\mathrm{to}\mathrm{}\mathrm{a}\mathrm{}\mathrm{circle}\mathrm{}\mathrm{are}\mathrm{}\mathrm{equal}. \\ \text{So},CR\mathit{=}CS=27\mathrm{c}\mathrm{m} \\ \therefore BR\mathit{=}\left(BC\mathit{-}CR\right) \\ \mathit{\Rightarrow }BR\mathit{=}\left(\mathit{38}\mathit{-}\mathit{27}\right)\mathrm{c}\mathrm{m} \\ \Rightarrow BR=11\mathrm{c}\mathrm{m} \\ \therefore BQ\mathit{=}BR=11\mathrm{c}\mathrm{m} \\ \mathrm{J}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{}OQ.\mathrm{} \\ \angle A=\angle Q=\angle O=\angle P=90°\mathrm{and}OP=OQ\mathit{}\left(\mathrm{radius}\right). \\ \text{So},PAQO\mathrm{i}\mathrm{s}\mathrm{}\mathrm{a}\mathrm{square}. \\ \therefore AQ\mathit{=}PQ=10\mathrm{c}\mathrm{m} \\ \mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e},\mathrm{}AB\mathit{=}\left(AQ\mathit{+}BQ\right) \\ \Rightarrow AB=\left(10+11\right)\mathrm{c}\mathrm{m} \\ \Rightarrow AB=21\mathrm{c}\mathrm{m} \end{gathered}$$