Question

Radius of circle is 10 cm. There are two chords of length 16 cm each. What will be the distance of these chords from the centre of the circle ?

Answer 1

Let the two chords be AB and CD.

AB = CD = 16 cm
Radius = AO = 10 cm
Let OE be the perpendicular drawn from the centre of the circle to the chord AB.
Perpendicular drawn from the centre of the circle to the chord bisects the chord.
So, AE = EB = 8 cm
In $△$AEO,
$$\begin{gathered} {\mathrm{EO}}^2+{\mathrm{AE}}^2={\mathrm{AO}}^2 \\ \Rightarrow {\mathrm{EO}}^2+8^2={10}^2 \\ \Rightarrow {\mathrm{EO}}^2+64=100 \\ \Rightarrow {\mathrm{EO}}^2=36 \\ \Rightarrow \mathrm{EO}=6\mathrm{cm} \end{gathered}$$
we know that congruent chords of a circle are equidistant from the centre.
So, EO = OF = 6 cm
Distance of these chords from the centre is 6 + 6 = 12 cm.