Question

In fig, $O$ is the centre of a circle $AB=16$ cm, $CD=14$ cm, seg $OM$ $\perp$ seg $AB$, seg $ON$ $\perp$ seg $CD$. If $OM=6$ cm, then length of seg $ON$ is $\sqrt{m}$cm. So, $m$ is

• A. $m=149{cm}^2$
• B. $m=51{cm}^2$
• C. $m=51$ cm
• D. $m=149$ cm

Answer 1

The correct option is B $m=51{cm}^2$

Given- $AB=16$ cm and $CD=14$ cm are the chordsof a circle with centre at $O$.

$OM(=6$ cm)$\perp AB$ at $M$ and $ON\perp CD$ at $N$.

To find out -

If the length of $ON=\sqrt{m}$ cm, then $m=?$

Solution-

We join $OC$and $OA$.

$\Delta OAM$ and $\Delta OCN$ are right ones, since $OM\perp AB$ at $M$ and $ON\perp CD$ at $N$.

Now $AM=\frac{1}{2}AB=\frac{1}{2}\times 16$ cm $=8$ cm and $CN=\frac{1}{2}CD=\frac{1}{2}\times 14$ cm $=7$ cm since the perpendicular from the centre of a circle to a chord bisects the latter.

So, in $\Delta OAM$, by Pythagoras theorem, we have

$OA=\sqrt{{OM}^2+{AM}^2}=\sqrt{6^2+8^2}$ cm $=10$ cm and it\ is the radius of the circle.

$\therefore OC=OA=10$ cm.

Again in $\Delta OCN$, by Pythagoras theorem, we have

$ON=\sqrt{{OC}^2-{CN}^2}=\sqrt{{10}^2-7^2}$ cm $=\sqrt{51}$ cm

But given that $ON=\sqrt{m}$.

$\therefore$ $m=51{cm}^2$