Question

$AB$ and $CD$ are two parallel chords of a circle such that $AB=10$ cm and $CD=24$ cm. If the chords are on opposite sides of the centre and the distance between them is $17$ cm. The radius of the circle is

• A. $26$ cm
• B. $39$ cm
• C. $6.5$ cm
• D. $13$ cm

Answer 1

The correct option is D $13$ cm

Given- $AB=10$ cm and $CD=24$ cm are two parallel\chords of a circle with centre $O$

To find out- the length of the radius of the given circle=?

Solution-

We join $OA$ and $OC$.

Also we drop perpendiculars $OM$ and $ON$ to $AB$ and $CD$ respectively.

So, $OA$ and $OC$ are the radii of the circle, i.e $OA=OC$.

$M$ and $N$ are the mid points of $AB$ and $CD$, since $ON\perp CD$ and $OM\perp AB$ and we know that the perpendicular, dropped from the center of a circle to any of its chords, bisects the latter.

$\therefore CD=2NC$ and $AB=2AM$

$\therefore NC=\frac{1}{2}CD=\frac{1}{2}\times 24$ cm $=12$ cm and $AM=\frac{1}{2}AB=\frac{1}{2}\times 10$ cm $=5$ cm .......(i)

Now the distance between the given chords $=MN$

Since $ON\perp CD$ and $OM\perp AB$ and $AB\parallel CD$, we have

$M,O$ and $N$ are collinear.

Let $ON=x$

$\therefore OM=17-x$

$\therefore$ in $\Delta OMA$ and $\Delta ONC$, we have

$OM=17-x,ON=x,\angle OMB=\angle OND$ ......(both are right angles)

and $OA=OC$ .....(radii of the same circle)

Therefore, by Pythagoras theorem, we have

${OA}^2={OM}^2+{AM}^2={(17-x)}^2+5^2$

and ${OC}^2=O{A}^2={NC}^2+{ON}^2={12}^2+x^2$

$\therefore {(17-x)}^2+5^2={12}^2+x^2$

$\Rightarrow 34x=170$

$\Rightarrow x=5$ cm

$\Rightarrow {OC}^2=O{A}^2={NC}^2+{ON}^2={12}^2+x^2$

$\Rightarrow {OC}^2={12}^2+5^2$

$\Rightarrow OC=13$ cm

So, the radius of the given circle is $13$ cm.