In a non-right-angled triangle ΔPQR, 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 ΔPQR equals 1, then which of the following options is/are correct ?

Engineering

Mathematics

Circumradius Inradius and exradius

Question

In a non-right-angled triangle ΔPQR, 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 ΔPQR equals 1, then which of the following options is/are correct ?

Area of $ΔSOE = \frac{\sqrt{3}}{12}$

Length of OE = $\frac{1}{6}$

Radius of incircle of $ΔPQR = \frac{\sqrt{3}}{2} (2 - \sqrt{3})$

Length of RS = $\frac{\sqrt{7}}{2}$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$\frac{sinP}{\sqrt{3}} = \frac{sinQ}{1} = \frac{1}{2 R} = \frac{1}{2}$

$\Rightarrow P = \frac{π}{3} or \frac{2 π}{3} and Q = \frac{π}{6} or \frac{5 π}{6}$

Since, p > q ⇒ P > Q

So, if P = $\frac{π}{3} and Q = \frac{π}{6} \Rightarrow R = \frac{π}{2} (not possible)$

Hence, P = $\frac{2 π}{3} and Q = R = \frac{π}{6}$

$r = \frac{Δ}{s} = \frac{\frac{1}{2} (1) (\sqrt{3}) (\frac{1}{2})}{(\frac{\sqrt{3} + 2}{2})} = \frac{\sqrt{3}}{2} (2 - \sqrt{3})$

Now, area of $ΔSEF = \frac{1}{4}$ area of $∆$ PQR

⇒ area of $ΔSOE = \frac{1}{3} area of ΔSEF = \frac{1}{12}$ area of $ΔPQR = \frac{1}{12} \cdot \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{48}$

$RS = \frac{1}{2} \sqrt{2 (3) + 2 (1) - 1} = \frac{\sqrt{7}}{2}$

$OE = \frac{PE}{3} = \frac{1}{3} \cdot \frac{1}{2} \sqrt{2 {(1)}^{2} + 2 {(1)}^{2} - 3} = \frac{1}{6}$