Question

In fig 4.39, O is the centre of a circle AB = 16cm, CD = 14cm, seg OM $\perp$ seg AB, seg ON $\perp$ seg CD. If OM =6cm, then find seg ON.

Answer 1

$$\begin{gathered} \mathrm{Join}\mathrm{AO}\mathrm{and}\mathrm{OC}. \\ \mathrm{AO}\mathrm{and}\mathrm{OC}\mathrm{are}\text{the}\mathrm{radi}\text{i of the circle.} \\ \therefore \mathrm{AO}=\mathrm{OC}. \\ \text{We}\mathrm{know}\mathrm{that}\mathrm{perpendicular}\mathrm{from}\mathrm{the}\mathrm{centre}\mathrm{bisects}\mathrm{the}\mathrm{chord}. \\ \therefore \mathrm{CN}=\mathrm{ND}=\frac{\mathrm{CD}}{2}=\frac{14}{2}=7\mathrm{cm} \\ \mathrm{And}\mathrm{AM}=\mathrm{MB}=\frac{\mathrm{AB}}{2}=\frac{16}{2}=8\mathrm{cm} \\ \mathrm{Now},\mathrm{in}\mathrm{right}\mathrm{triangle}\mathrm{AMO}, \\ \mathrm{AO}=\sqrt{{\mathrm{AM}}^2+{\mathrm{MO}}^2} \\ =\sqrt{8^2+6^2} \\ =\sqrt{100} \\ =10\mathrm{cm} \\ \therefore \mathrm{OC}=10\mathrm{cm} \\ \mathrm{Now},\mathrm{in}\mathrm{right}\mathrm{triangle}\mathrm{CNO}, \\ \mathrm{ON}=\sqrt{{\mathrm{OC}}^2-{\mathrm{CN}}^2} \\ =\sqrt{{10}^2-7^2} \\ =\sqrt{51}\mathrm{cm} \\ \therefore \mathrm{ON}=\sqrt{51}\mathrm{cm}. \end{gathered}$$