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 from R meets the side PQ at S, the perpendicular from P meets the side QR at E, and RS and PE intersect at O. If $p=\sqrt{3},q=1,$ and the radius of the circumcircle of the $△PQR$ equals $1$, 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. Length of OE = $\frac{1}{6}$
• D. Radius of incircle of $△PQR=\frac{\sqrt{3}}{2}(2-\sqrt{3})$

Answer 1

Answer: (A), (C), (D)

The correct options are
A Length of RS = $\frac{\sqrt{7}}{2}$
C Length of OE = $\frac{1}{6}$
D Radius of incircle of $△PQR=\frac{\sqrt{3}}{2}(2-\sqrt{3})$

Using Sine law,
$\Rightarrow \frac{PR}{\sin Q}=\frac{QR}{\sin P}=2R=2$
$PR=q=1,p=QR=\sqrt{3}$
$\sin P=\frac{\sqrt{3}}{2}\Rightarrow \angle P=\frac{\pi }{3}or\frac{2\pi }{3}$
$\sin Q=\frac{1}{2}\Rightarrow \angle Q=\frac{\pi }{6}or\frac{5\pi }{6}$
Since $p>q\Rightarrow \angle P>\angle Q$
Hence, $\angle P=\frac{2\pi }{3}$ and $\angle Q=\angle R=\pi /6$
$PQ=2R\sin R=\mathrm{2.1.}\sin \frac{\pi }{6}=1$
Using Apollonius' theorem
$2\left(\right(RS{)}^2+\left(PS{)}^2\right)=(PR{)}^2+(RQ{)}^2$
$2\left(\right(RS{)}^2+\left(\frac{1}{2}{)}^2\right)=(1{)}^2+(\sqrt{3}{)}^2$
Therefore, Length of $RS=\frac{\sqrt{7}}{2}$

Radius of incircle $r=4R\sin \frac{P}{2}.\sin \frac{Q}{2}.\sin \frac{R}{2}=\mathrm{4.1.}\frac{\sqrt{3}}{2}{\sin }^2{15}^o$
$r=\frac{\sqrt{3}}{2}(2-\sqrt{3})$

From Fig. we can observe that,
$(PE{)}^2+(RE{)}^2=(PR{)}^2$
$\Rightarrow (PE{)}^2=1-\frac{3}{4}$
$PE=\frac{1}{2}$
From $△PQR$, using property
$\frac{OE}{PE}=\frac{1}{3}\Rightarrow OE=\frac{1}{6}$

Now,
area of $△SEF=\frac{1}{4}\times$ area of $△PQR$
$=\frac{1}{4}.\frac{1}{2}.\sqrt{3}.\frac{1}{2}=\frac{\sqrt{3}}{16}$

area of $△SOE=\frac{1}{3}\times$ area of $△SEF=\frac{1}{12}\times$ area of $△PQR$
$=\frac{1}{12}.\frac{1}{2}.\sqrt{3}.\frac{1}{2}=\frac{\sqrt{3}}{48}$