Question

Distance between equal chords, $AB$ and $CD$ of a circle with centre, $O$ and radius, $5cm$ is $8cm$. Find the length of the the chords.

Answer 1

Consider the figure:

We know that equal chords are equidistant from the center.

Since $AB=CD$, we must have $OE=OF$.

Now, $EF=8cm$ [Given]

$OE+OF=EF=8cm$

$\Rightarrow 2OE=8$

$\Rightarrow OE=4cm=OF$

In $△OCF$, applying Pythagoras theorem,

$O{C}^2=O{F}^2+C{F}^2$

$\Rightarrow 5^2=4^2+C{F}^2$

$\Rightarrow C{F}^2=5^2-4^2=25-16=9$

$\Rightarrow CF=3cm$

We know that perpendicular from the centre to the chord bisects the chord.

So, $CF=FD=3cm$

$\Rightarrow CD=2\times CF=2\times 3=6cm$

So, length of the chords is $6cm$.