In a non-right-angled triangle Δ P Q R , let p , q , r denote the lengths of the sides opposite to the…

1. Length of $RS=\frac{\sqrt{7}}{2}$
2. Area of $\mathrm{\Delta }SOE=\frac{\sqrt{3}}{12}$
3. Radius of incircle of $\mathrm{\Delta }PRQ=\frac{\sqrt{3}}{2}\left(2-\sqrt{3}\right)$
4. Length of $OE=\frac{1}{6}$

Solution

By sine rule in $PQR$ $\frac{p}{\mathrm{s}\mathrm{i}\mathrm{n}P}=\frac{q}{\mathrm{s}\mathrm{i}\mathrm{n}Q}=\frac{r}{\mathrm{s}\mathrm{i}\mathrm{n}R}=2R$ $\frac{\sqrt{3}}{\mathrm{s}\mathrm{i}\mathrm{n}P}=\frac{1}{\mathrm{s}\mathrm{i}\mathrm{n}Q}=\frac{1}{\mathrm{s}\mathrm{i}\mathrm{n}R}=2\left(1\right)$ $\mathrm{s}\mathrm{i}\mathrm{n}P=\frac{\sqrt{3}}{2}$ and $\mathrm{s}\mathrm{i}\mathrm{n}Q=\mathrm{s}\mathrm{i}\mathrm{n}R=\frac{1}{2}$ $P={60}^{°}$ or ${120}^{°}, Q={30}^{°}$ and ${150}^{°}$ $\Rightarrow$ As $PQR$ is not a right angle triangle Only possibility is $P={120}^{°},\mathrm{ }Q=R={30}^{°}$ $\Rightarrow PQR$ is an isosceles triangle, hence, $PE$ is also a median and $‘O’$ will be centroid. ( $∆PQE$ and $∆PRE$ are congruent and $E$ is mid point of $QR$ ) $\Rightarrow r=q=1, P=\sqrt{3}$ Option $\left(A\right):$ Length of median from $R$ $RS=\frac{1}{2}\sqrt{2p^2+2q^2-r^2}=\frac{1}{2}\sqrt{2{\left(\sqrt{3}\right)}^2+2\left(1\right)-1}=\frac{\sqrt{7}}{2}$ Option $\left(B\right):$ Now, area of $\Delta SEF=\frac{1}{4}\Delta PQR$   ...$\left(\mathrm{i}\right)$ and area of $\Delta SOE=\frac{1}{3}\Delta SEF$ ...$\left(\mathrm{ii}\right)$ from $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$ $\Delta SOE=\frac{1}{12}\Delta PQR$ ...$\left(\mathrm{iii}\right)$ $\Delta PQR=\frac{1}{2}pq \mathrm{s}\mathrm{i}\mathrm{n}R=\frac{1}{2}\times \sqrt{3}\times 1\times \frac{1}{2}=\frac{\sqrt{3}}{4}$ $\Delta SOE=\frac{1}{12}\Delta PQR=\frac{\sqrt{3}}{48}$ Option $\left(D\right):$ Length of $OE$ As $O$ is centroid $OE$ will be $\frac{1}{3}$ of $PE$ $PE=\frac{1}{2}\sqrt{2q^2+2r^2-p^2}=\frac{1}{2}\sqrt{2+2-3}$ $=\frac{1}{2}$ $OE=\frac{1}{3} PE=\frac{1}{6}$ Option $\left(C\right):$ Radius of incircle $=\frac{\Delta }{S}\mathrm{ } \left(\mathrm{i}.\mathrm{e}.\frac{\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}}{\mathrm{s}\mathrm{e}\mathrm{m}\mathrm{i}-\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}\right)$ $=\frac{\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\left(\frac{1+1+\sqrt{3}}{2}\right)}$ $=\frac{\sqrt{3}}{2\left(2+\sqrt{3}\right)}$ $=\frac{\sqrt{3}}{2}\left(2-\sqrt{3}\right)$

Asked in: JEE Advanced 2019 (Paper 1)

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