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Q.

A circle is drawn in a sector of a larger circle of radius r, as shown in figure. The smaller circle is tangent to the two
bounding radii and the arc of the sector. The radius of the smaller circle is

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a

$\frac{r}{2}$

b

$\frac{r}{3}$

c

$\frac{2 \sqrt{3} r}{5}$

d

$\frac{r}{\sqrt{2}}$

answer is B.

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Detailed Solution

Let the radius of smaller circle be x. Here,

$\begin{matrix} and OP = xcosec 30^{\circ} \\ \Rightarrow OQ = r = x + xcosec 30^{\circ} \\ x = \frac{r}{3} \end{matrix}$

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