Question

Two parallel chords in a circle are $10cm$ and $24cm$ long. If the radius of the circle is $13cm$, find the distance between the chords if thay lie on the opposite sides of the center

Answer 1

Let $AB=10cm$ and $CD=24cm$ be the chords.

Then $OA=OB=OC=OD=13cm$ [radius]

Since the perpendicular from the centre $O$ to the chord bisects the chord.
Therefore $E$ and $F$ will be the midpoints of $AB$ and $CD$ respectively.

$AE=EB=\frac{1}{2}\times AB$

$=\frac{1}{2}\times 10=5cm$

And $CF=FD=\frac{1}{2}\times CD$

$=\frac{1}{2}\times 24=12cm$

In $△AEO,$

$O{A}^2=A{E}^2+O{E}^2$ [By Pythagoras theorem]

$O{E}^2=(13{)}^2-(5{)}^2$

$OE=\sqrt{144}=12cm$

In $△CFO,$

$O{C}^2=C{F}^2+O{F}^2$ [By Pythagoras theorem]

$O{F}^2=(13{)}^2-(12{)}^2$

$OF=\sqrt{25}=5cm$

So, $EF=12+5=17cm$

$\therefore$ Distance between chords if they lie on opposite sides centers $=17cm$