Question

In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the centre. Find the lengths of the chords.

Answer 1

Let the two chords be AB and CD.

Radius = AO = 13 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
In $△$AEO,
$$\begin{gathered} {\mathrm{EO}}^2+{\mathrm{AE}}^2={\mathrm{AO}}^2 \\ \Rightarrow 5^2+{\mathrm{AE}}^2={13}^2 \\ \Rightarrow {\mathrm{AE}}^2+25=169 \\ \Rightarrow {\mathrm{AE}}^2=144 \\ \Rightarrow \mathrm{AE}=12\mathrm{cm} \end{gathered}$$
And AE = EB = 12 + 12 = 24 cm
Thus, AB = CD = 24 cm as chords of a circle equidistant from the centre are congruent.