Question

In a non-right-angled triangle $△PQR$, Let $p,q,r$ denote the lengths of the sides opposite to the angles at $P,Q,R$ respectively. The median form $R$ meets the side $PQ$ at $S$, the perpendicular $4$ from $P$ meets the side $QR$ at $E$, and $RS$ and $PE$ intersect at $O$. If $p=3,q=1,$ and the radius of the circumcircle of the $△PQR$ equals $1$, then which of the following options is/are correct?

• A. Length of $RS=\frac{\sqrt{7}}{2}$
• B. Area of $△SOE=\frac{\sqrt{3}}{12}$
• C. Radius of incircle of $△PQR=\frac{\sqrt{3}}{2}(2-\sqrt{3})$
• D. Length of $OE=\frac{1}{6}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Length of $RS=\frac{\sqrt{7}}{2}$
B Area of $△SOE=\frac{\sqrt{3}}{12}$
C Radius of incircle of $△PQR=\frac{\sqrt{3}}{2}(2-\sqrt{3})$

$\frac{p}{\sin P}=\frac{q}{\sin Q}=2\left(1\right)\Rightarrow \sin P=\frac{\sqrt{3}}{2},\sin Q=\frac{1}{2}$
$\Rightarrow \angle P={60}^{\circ }$ or ${120}^{\circ }$ and $\angle Q={30}^{\circ }$ or ${150}^{\circ }$
because $\angle P+\angle Q$ must be less than ${180}^{\circ }$ but not equal to ${90}^{\circ }$
$\angle P={120}^{\circ }$ and $\angle Q={30}^{\circ }$ and $\angle R={30}^{\circ }\frac{r}{\sin R}=2\Rightarrow r=1$
Now length of median $RS=\frac{1}{2}\sqrt{2p^2+2q^2-r^2}=\frac{1}{2}\sqrt{6+2-1}=\frac{\sqrt{7}}{2}\Rightarrow$ option (A) is correct
Inradius $=\frac{2△}{p+q+r}=\frac{\frac{2pqr}{4\times \left(1\right)}}{p+q+r}=\frac{1}{2}\left(\frac{1\times 1\times \sqrt{3}}{1+1+\sqrt{3}}\right)=\frac{\sqrt{3}}{2}\left(\frac{2-\sqrt{3}}{1}\right)\Rightarrow$ option (C) is correct
$\Rightarrow \frac{1}{2}\times \sqrt{3}\times PE=\frac{pqr}{4\left(1\right)}$ (equal area of $△)\Rightarrow PE=\frac{1\times 1\times \sqrt{3}}{4}\times \frac{2}{\sqrt{3}}=\frac{1}{2}$
$\Rightarrow OE=\frac{2\left(Areaof△OQR\right)}{QR}=\frac{2\times \frac{1}{3}(\frac{1}{2}.1.\sqrt{3}\sin {30}^{\circ })}{\sqrt{3}}=\frac{1}{6}$.