QJEEAdv19Ma12a1: JEE Advanced 2019 Mathematics - Class 11 JEEAdv19Ma12a1 Solution: In a non-right-angled triangle PQR, let p, q, r denote the…

Let a non-right angled $\Delta$ PQR.

Now, by sine rule



${P \over {\sin P}} = {q \over {\sin Q}} = {r \over {\sin R}} = 2 \times$ circumradius

$\Rightarrow {{\sqrt 3 } \over {\sin P}} = {1 \over {\sin Q}} = {r \over {\sin R}} = 2 \times 1$

[circumradius = 1 unit]

$\Rightarrow$ ${\sin P}$ = ${{\sqrt 3 } \over 2}$ and sinQ = ${1 \over 2}$

$\Rightarrow P = 120^\circ$ and $\Rightarrow Q = 30^\circ$

( $\because$ $\Delta$ PQR is non-right angled triangle)

So, R = 30 $^\circ$

$\Rightarrow$ r = 1, so $\Delta$ PQR is an isosceles triangle. And, since RS and PE are the median of $\Delta$ PQR, so 'O' is centroid of the $\Delta$ PQR.

Now,

Option (a),

From Apollonius theorem,

$2(P{E^2} + Q{E^2}) = P{Q^2} + P{R^2}$

$\Rightarrow 2\left( {P{E^2} + {3 \over 4}} \right) = 1 + 1$

$\Rightarrow P{E^2} = 1 - {3 \over 4} \Rightarrow P{E^2} = {1 \over 4}$

$\Rightarrow PE = {1 \over 2}$ units

and $OE = {1 \over 3}PE = {1 \over 6}$ units

[ $\because$ O divides PE is 2 : 1]

Option (b),

Again from Apollonius theorem,

2 (PS² + RS²) = PR² + QR²

$\Rightarrow 2\left( {{1 \over 4} + R{S^2}} \right) = 1 + 3$

$\Rightarrow R{S^2} = 2 - {1 \over 4} \Rightarrow R{S^2} = {7 \over 4}$

$\Rightarrow RS = {{\sqrt 7 } \over 2}$ units

Option (c),

Area of $\Delta$ SOE = ${1 \over 2}(OE)\,(ST)$

= ${1 \over 2} \times {1 \over 6}[(PS)sin60^\circ ]$

= ${1 \over {12}} \times {1 \over 2} \times {{\sqrt 3 } \over 2}$

= ${{\sqrt 3 } \over {48}}$ square units

Option (d),

$\because$ Inradius of

$\Delta PQR = {\Delta \over s} = {{{1 \over 2}pq\sin R} \over {{1 \over 2}(p + q + r)}} = {{{1 \over 2}(\sqrt 2 )(1){1 \over 2}} \over {{1 \over 2}(\sqrt 3 + 1 + 1)}}$

$= {{\sqrt 3 } \over 2}(2 - \sqrt 3 )$ units

Hence, options (a), (b) and (d) are correct.