Question

In a circle of radius 17cm two parallel chords of the length 30cm and 16cm respectively are drawn on the opposite side of the centre then find the distance between them.

Answer 1

$$\begin{gathered} \text{Given:}\mathrm{PQ}\hspace{0.17em}=30\mathrm{cm}\mathrm{and}\mathrm{RS}=16\mathrm{cm} \\ D\mathrm{raw}\mathrm{a}\mathrm{line}\mathrm{segment}\mathrm{MN},so\mathrm{that}\mathrm{it}\mathrm{is}\mathrm{perpendicular}\mathrm{on}\mathrm{both}\mathrm{PQ}\mathrm{and}\mathrm{RS}\mathrm{and}\mathrm{it}\mathrm{passes}\mathrm{through}\text{the}\mathrm{centre}\mathrm{O}. \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{perpendicular}\mathrm{from}\mathrm{the}\mathrm{centre}\mathrm{on}\text{a}\mathrm{chord}\mathrm{divides}\mathrm{the}\mathrm{chord}. \\ \therefore \mathrm{PM}\hspace{0.17em}=\mathrm{MQ}=15\mathrm{cm}\mathrm{and}\mathrm{RN}=\mathrm{NS}=8\mathrm{cm}. \\ \mathrm{Now},\mathrm{in}\mathrm{right}\mathrm{triangle}\mathrm{OMQ}, \\ \mathrm{OM}=\sqrt{{\mathrm{OQ}}^2-{\mathrm{MQ}}^2} \\ =\sqrt{{17}^2-{15}^2} \\ =\sqrt{289-225} \\ =\sqrt{64} \\ =8\mathrm{cm} \\ \mathrm{Similarly}\mathrm{in}\mathrm{right}\mathrm{triangle}\mathrm{ONS}, \\ \mathrm{ON}=\sqrt{{\mathrm{OS}}^2-{\mathrm{SN}}^2} \\ =\sqrt{{17}^2-8^2} \\ =\sqrt{289-64} \\ =\sqrt{225} \\ =15\mathrm{cm} \\ \mathrm{Now},\mathrm{MN}=\mathrm{ON}+\mathrm{OM} \\ =15\mathrm{cm}+8\mathrm{cm}=23\mathrm{cm} \\ \therefore \mathrm{The}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{chords}=23\mathrm{cm}. \end{gathered}$$