Question

Assertion :Statement-l: From a point $P$ on the circle with centre $O$ the chord $PA=8$ cm is drawn. The radius of the circle is $24$ cm. Let $PB$ be drawn parallel to $OA$. Suppose $BO$ extended meet $PA$ extended at $M$. The length of$MA$ is $9$ cm. Reason: Statement-2: $OA$ is a radius of a circle with centre at $O.R$ is a point on $OA$ through which a chord $CD$ perpendicular to $OA$ is drawn. Let a chord through A meet the chord $CD$ at $M$ and the circle at $B$. Also $OS$ is perpendicular from $O$ on chord $AB$. The radius of the circle is $18$ cm. $R$ is the mid point of $AO$ and $AM/MB=\frac{1}{2},$ The length of $OS$ is $9$ cm.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

In statement-$1$, let $AN=x,ON=r-x$ where

$r=24$ $BQ=QP=ON=r-x$ From $\Delta OQB,(OQ{)}^2=(OB{)}^2-(BQ{)}^2=r^2-(r-x{)}^2$

From $\Delta PNA,(PN{)}^2=(PA{)}^2-(AN{)}^2=64-x^2$

$OQ=PN\Rightarrow r^2-(r-x{)}^2=64-x^2$

$\Rightarrow 2rx=64\Rightarrow x=\frac{64}{2\times 24}=\frac{4}{3}$

From $\Delta MPB,$ since $BP$ is parallel to $OA$
$\frac{MP}{MA}=\frac{BP}{OA}$

$\Rightarrow \frac{MP+AP}{MA}=\frac{2(r-x)}{r}$

$\Rightarrow MA=\frac{8r}{r-2x}=\frac{8\times 24}{24-2\times \frac{4}{3}}=\frac{8\times 24\times 3}{72-8}=9$

So statement-$1$ is true. In statement-$2$

From similar triangles $ARM$ and $ASO$

$\frac{AM}{AO}=\frac{AR}{AS}\Rightarrow AR\times AO=AM\times AS$

$\frac{AM}{MB}=\frac{1}{2}\Rightarrow AM=\frac{1}{3}AB$

Also AS $=\left(\frac{1}{2}\right)AB$ and $AR=\left(\frac{1}{2}\right)AO$
so$AR\times AO=AM\times AS$

$\Rightarrow \frac{1}{2}(AO{)}^2=\frac{1}{6}(AB{)}^2$

$\Rightarrow (AB{)}^2=3\times (18{)}^2\Rightarrow AB=18\sqrt{3}$

$\Rightarrow AS=\left(\frac{1}{2}\right)AB=9\sqrt{3}$

$(OS{)}^2=(OA{)}^2-(AS{)}^2=(18{)}^2-9^2\cdot 3=9^2$

$\Rightarrow OS=9$ cm and thus statement-$2$ is also True. But statement-$2$ is not a, correct explanation for statement-$1$