Question

In a circle of radius 10 cm, find the length of a chord which is at a distance of 5 cm from its centre.

Answer 1

We know that the perpendicular drawn from the centre of a circle bisects the chord of the circle.
So, AC = BC
$$\begin{gathered} \mathrm{Now},\text{in}∆AOC, \\ A{O}^2=A{C}^2+O{C}^2 \\ \Rightarrow AC=\sqrt{A{O}^2-O{C}^2} \\ =\sqrt{{10}^2-5^2} \\ =\sqrt{100-25} \\ =\sqrt{75} \\ =5\sqrt{3} \\ =5\times 1.732 \\ =8.66\mathrm{cm} \\ \mathrm{AB}=2\times 8.66=17.32\mathrm{cm} \end{gathered}$$