JEE Advance - Mathematics (2019 - Paper 1 Offline - No. 7)
In a non-right-angled triangle $$\Delta $$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 $$\Delta $$PQR equals 1, then which of the following options is/are correct?
Length of OE = $${1 \over 6}$$
Length of RS = $${{\sqrt 7 } \over 2}$$
Area of $$\Delta $$SOE = $${{\sqrt 3 } \over {12}}$$
Radius of incircle of $$\Delta $$PQR = $${{\sqrt 3 } \over {2}}$$($${2 - \sqrt 3 }$$)
Explanation
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.
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.
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