Question

In fig 4.41 chord AB || chord CD of a circle with centre O and radius 5 cm such that AB = 6cm and CD = 8cm, if OP $\perp$ AB, OQ $\perp$ CD, Determine PQ.

Answer 1

$$\begin{gathered} \mathrm{Join}\mathrm{OA}\mathrm{and}\mathrm{OC}. \\ \therefore \mathrm{Radius}=5\mathrm{cm} \\ \therefore \mathrm{OA}\hspace{0.17em}=\mathrm{OC}\hspace{0.17em}=5\mathrm{cm} \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{perpendicular}\mathrm{from}\mathrm{the}\mathrm{centre}\mathrm{of}\text{a}\mathrm{circle}\mathrm{on}\text{a}\mathrm{chord}\mathrm{bisects}\mathrm{the}\mathrm{chord}. \\ \therefore \mathrm{AP}=\mathrm{BP}=\frac{\mathrm{AB}}{2}=\frac{6}{2}=3\mathrm{cm} \\ \mathrm{And}\mathrm{CQ}\hspace{0.17em}=\mathrm{DQ}=\frac{\mathrm{CD}}{2}=\frac{8}{2}=4\mathrm{cm} \\ \mathrm{In}\mathrm{right}△\mathrm{OAP}, \\ \mathrm{OP}=\sqrt{{\mathrm{OA}}^2-{\mathrm{AP}}^2} \\ =\sqrt{5^2-3^2} \\ =\sqrt{25-9} \\ =\sqrt{16} \\ =4\mathrm{cm} \\ \mathrm{And}\mathrm{in}\mathrm{right}△\mathrm{OCQ}, \\ \mathrm{OQ}=\sqrt{{\mathrm{OC}}^2-{\mathrm{CQ}}^2} \\ =\sqrt{5^2-4^2} \\ =\sqrt{25-16} \\ =\sqrt{9} \\ =3\mathrm{cm} \\ \mathrm{From}\text{the}\mathrm{figure}, \\ \mathrm{PQ}=\mathrm{OP}-\mathrm{OQ}=4\mathrm{cm}-3\mathrm{cm}=1\mathrm{cm} \\ \therefore \mathrm{PQ}=1\mathrm{cm} \end{gathered}$$