Question

In the given figure, O is the centre of a circle, such that OA = 5 cm, AB = 8 cm and OD ⊥ AB, meeting AB at C. Find the length of CD.

Answer 1

$$\begin{gathered} OD\perp AB. \\ \text{W}e\mathrm{know}\mathrm{that}\text{the}\mathrm{perpendicular}\mathrm{drawn}\mathrm{from}\mathrm{the} \\ \mathrm{centre}\mathrm{of}\text{a}\mathrm{circle}\mathrm{to}\text{a}\mathrm{chord}\mathrm{bisects}\mathrm{the}\mathrm{chord}. \\ \mathrm{Hence},OD\mathrm{will}\mathrm{bisect}AB. \\ So,AC=\frac{1}{2}AB=\frac{1}{2}\times 8=4\mathrm{cm} \\ ∆OCA\mathit{}\mathrm{is}\mathrm{right}-\mathrm{angled}\mathrm{triangle}. \\ \mathrm{So},\mathrm{by}\text{P}\mathrm{ythagoras}\text{'}\mathrm{theorem}: \\ OC=\sqrt{O{A}^{\mathit{2}}-A{C}^2} \\ =\sqrt{5^{\mathit{2}}-4^2} \\ =\sqrt{25-16} \\ =\sqrt{9} \\ =3\mathrm{cm} \\ \mathrm{and} \\ CD=OD-OC\mathit{} \\ =5-3 \\ =2\mathrm{cm} \end{gathered}$$