Question

Two chords PQ and MN of length 11 cm and 5 cm respectively of a circle are parallel to each other and are on the oposite sides of its centre. If the distance between chord MN and chord PQ is 6cm find the radius of the circle.

Answer 1

Given : PQ = 11 cm, MN = 5 cm and KL = 6 cm
Let OK = x cm. Then, OL = (6 $-$ x) cm.
$$\begin{gathered} \mathrm{KQ}=\frac{\mathrm{PQ}}{2} \\ \therefore \mathrm{KQ}=\frac{11}{2}\mathrm{cm}(OK\mathrm{is}\mathrm{the}\mathrm{perpendicular}\mathrm{form}\mathrm{the}\mathrm{centre}\mathrm{to}\mathrm{the}\mathrm{chord}PQ) \\ \mathrm{Now}\mathrm{in}\mathrm{right}\mathrm{triangle}\mathrm{OKQ}, \\ {\mathrm{OQ}}^2={\mathrm{OK}}^2+{\mathrm{KQ}}^2 \\ \Rightarrow {\mathrm{OQ}}^2=x^2+{\left(\frac{11}{2}\right)}^2 \\ \Rightarrow {\mathrm{OQ}}^2=x^2+\frac{121}{4}------\left(1\right) \\ \mathrm{LN}=\frac{\mathrm{MN}}{2}(OL\mathrm{is}\mathrm{the}\mathrm{perpendicular}\mathrm{form}\mathrm{the}\mathrm{centre}\mathrm{to}\mathrm{the}\mathrm{chord}MN) \\ \therefore \mathrm{LN}=\frac{5}{2}\mathrm{cm} \\ \mathrm{Now}\mathrm{in}\mathrm{right}\mathrm{triangle}\mathrm{OLN}, \\ {\mathrm{ON}}^2={\mathrm{OL}}^2+{\mathrm{LN}}^2 \\ \Rightarrow {\mathrm{ON}}^2={\left(6-x\right)}^2+{\left(\frac{5}{2}\right)}^2 \\ \Rightarrow {\mathrm{ON}}^2={\left(6-x\right)}^2+\frac{25}{4}------\left(2\right) \\ \mathrm{Now}, \\ \mathrm{OQ}=\mathrm{ON}\left(\mathrm{radii}\mathrm{of}\mathrm{same}\mathrm{circle}\right) \\ \Rightarrow {\mathrm{OQ}}^2={\mathrm{ON}}^2 \\ \Rightarrow x^2+\frac{121}{4}={\left(6-x\right)}^2+\frac{25}{4} \\ \Rightarrow x^2+\frac{121}{4}=36+x^2-12x+\frac{25}{4} \\ \Rightarrow \frac{121}{4}-\frac{25}{4}=36-12x \\ \Rightarrow 24=36-12x \\ \Rightarrow 12x=12 \\ \Rightarrow x=1 \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{in}\mathrm{equation}\left(1\right)\mathrm{we}\mathrm{get}: \\ {\mathrm{OQ}}^2=1^2+\frac{121}{4} \\ \Rightarrow {\mathrm{OQ}}^2=\frac{125}{4} \\ \Rightarrow \mathrm{OQ}=\frac{5\sqrt{5}}{2} \\ \therefore \text{R}\mathrm{adius}=\frac{5\sqrt{5}}{2}\mathrm{cm} \end{gathered}$$