Question

HI and JK are two parallel chords of a circle which are on opposite sides of the centre such that HI = 8cm, JK = 6cm and the distance between them is 7cm then find the radius of the circle.

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{HI}\mathrm{and}\mathrm{JK}\mathrm{are}\mathrm{two}\mathrm{par}a\mathrm{llel}\mathrm{chords}\mathrm{of}\mathrm{length}8\mathrm{cm}\mathrm{and}6\mathrm{cm},\mathrm{respectively}. \\ \mathrm{The}\mathrm{distance}\mathrm{between}\mathrm{them}\mathrm{is}7\mathrm{cm},\mathrm{i}.\mathrm{e}.\mathrm{MN}=7\mathrm{cm}. \\ \mathrm{Let}\mathrm{OM}=\mathit{}x\mathrm{cm}. \\ \mathrm{Then},\mathrm{ON}\hspace{0.17em}=(7-x)\mathrm{cm}. \\ \mathrm{Now},\mathrm{in}\mathrm{right}\mathrm{triangle}\mathrm{OMI}, \\ {\mathrm{OI}}^2={\mathrm{OM}}^2+{\mathrm{MI}}^2 \\ \Rightarrow {\mathrm{OI}}^2=x^2+4^2 \\ \Rightarrow {\mathrm{OI}}^2=x^2+16----\left(1\right) \\ \mathrm{And},\mathrm{in}\mathrm{right}\mathrm{triangle}\mathrm{ONK}, \\ {\mathrm{OK}}^2={\mathrm{ON}}^2+{\mathrm{NK}}^2 \\ \Rightarrow {\mathrm{OK}}^2={\left(7-x\right)}^2+3^2 \\ \Rightarrow {\mathrm{OK}}^2={\left(7-x\right)}^2+9----\left(2\right) \\ \mathrm{Now},\mathrm{OI}\mathrm{and}\mathrm{OK}\mathrm{are}\mathrm{rad}\text{ii of the circle}. \\ \mathrm{So},\mathrm{length}\mathrm{of}\mathrm{both}\mathrm{will}\mathrm{be}\mathrm{same}, \\ \mathrm{OI}=\mathrm{OK} \\ \Rightarrow \mathrm{OI}{}^2={\mathrm{OK}}^2 \\ \Rightarrow x^2+16={\left(7-x\right)}^2+9 \\ \Rightarrow x^2+16=49-14x+x^2+9 \\ \Rightarrow 14x\mathit{}=42 \\ \Rightarrow x=3 \\ \mathrm{Using}\left(1\right),\mathrm{we}\mathrm{get}: \\ {\mathrm{OI}}^2=3^2+16=25 \\ \therefore \mathrm{OI}=5\mathrm{cm} \\ \mathrm{Therefore},\mathrm{radius}\mathrm{of}\text{the}\mathrm{circle}=5\mathrm{cm}. \end{gathered}$$