Chapter 10.1, Problem 11PSB

To determine

Distance from centre of circle to a chord.

Answer to Problem 11PSB

  $8\text{m}$

Explanation of Solution

Given information : $\text{Chord is 30 m long and diameter of the circle is 34 cm}\text{.}$

  $\begin{array}{l}\text{AB is diameter and CD is chord; O is centre}\text{.}\\ \text{Join OC; AO = OB = OC = 17 m}\\ \because \text{AO = }\frac{\text{AB}}{\text{2}}\text{ and CE = }\frac{\text{CD}}{\text{2}}=\frac{30}{2}=15\text{m}\\ \text{In triangle OCE,}\\ \text{Using Pythagoras Theorem,}\\ {\text{OC}}^{\text{2}}={\text{OE}}^2+{\text{EC}}^2\\ {17}^2=O{E}^2+{15}^2\\ O{E}^2={17}^2-{15}^2=289-225=64\\ OE=8\text{m}\\ \text{Distance between chord and diameter is 8m}\text{.}\end{array}$