Question

In the given figure, ΔABC is right-angled at B, such that BC = 6 cm and AB = 8 cm. A circle with centre O has been inscribed in the triangle. OP ⊥ AB, OQ ⊥ BC and OR ⊥ AC.
If OP = OQ = OR = x cm, then x = ?


(a) 2 cm
(b) 2.5 cm
(c) 3 cm
(d) 3.5 cm

Answer 1

(a) 2 cm
$$\begin{gathered} \mathrm{Given},AB=8\mathrm{cm},BC=6\mathrm{}\mathrm{cm} \\ \mathrm{Now},\text{in}∆ABC: \\ A{C}^2=A{B}^2+B{C}^2 \\ \Rightarrow A{C}^2=\left(8^2+6^2\right) \\ \Rightarrow A{C}^2=\left(64+36\right) \\ \Rightarrow A{C}^2=100 \\ \Rightarrow AC=\sqrt{100} \\ \Rightarrow AC=10\mathrm{cm} \\ PBQO\mathrm{is}\mathrm{a}\mathrm{square}. \\ CR=CQ\left(\mathrm{Since}\mathrm{}\mathrm{the}\mathrm{}\mathrm{length}s\mathrm{}\mathrm{of}\mathrm{}\mathrm{tangents}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{from}\mathrm{an}\mathrm{external}\mathrm{}\mathrm{point}\mathrm{}\mathrm{are}\mathrm{}\mathrm{equal}\right) \\ \therefore CQ=\left(BC-BQ\right)=\left(6-x\right)\mathrm{cm} \\ \mathrm{Similarly},AR=AP=\left(AB-BP\right)=\left(8-x\right)\mathrm{cm} \\ \therefore AC=\left(AR+CR\right)=\left[\left(8-x\right)+\left(6-x\right)\right]\mathrm{cm} \\ \Rightarrow 10\mathrm{}=\left(14-2x\right)\mathrm{cm} \\ \Rightarrow 2x=4 \\ \Rightarrow x=2\mathrm{cm} \\ \therefore \mathrm{The}\mathrm{}\mathrm{radius}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{circle}\mathrm{}\mathrm{is}\mathrm{}2\mathrm{cm}. \end{gathered}$$