Question

Consider a semicircle of radius $1$ unit constructed on the diameter $AB$, and let $O$ be its centre. Let $C$ be a point on $AO$ such that $AC:CO=2:1$. Draw $CD$ perpendicular to $AO$ with $D$ on the semicircle. Draw $OE$ perpendicular to $AD$ with $E$ on $AD$. Let $OE$ and $CD$ intersects at $H$. Then $DH$ equals

• A. $\frac{1}{\sqrt{5}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{\sqrt{5}-1}{2}$

Answer 1

The correct option is C $\frac{1}{\sqrt{2}}$


As $OD=OA$ so $E$ is the midpoint of $AD$.
$AC:CO=2:1\Rightarrow AC=\frac{2}{3},CO=\frac{1}{3}$
In $△CDO\cos 2\theta =\frac{CO}{DO}\Rightarrow \cos 2\theta =\frac{1}{3}\Rightarrow 1-2{\sin }^2\theta =\frac{1}{3}\Rightarrow \sin \theta =\frac{1}{\sqrt{3}};\cos \theta =\sqrt{\frac{2}{3}}$
In $△EOD$,
$\sin \theta =\frac{ED}{1}\Rightarrow ED=\frac{1}{\sqrt{3}}$
In $△EHD$,
$\cos \theta =\frac{ED}{DH}\Rightarrow \sqrt{\frac{2}{3}}=\frac{1}{\sqrt{3}DH}\Rightarrow DH=\frac{1}{\sqrt{2}}$