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JEE Advanced 2019 Paper 1
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© examsnet.com
Question : 47
Total: 54
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, theperpendicular 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 ΔPQR equals 1, then which of the following options is/are correct?
Length of
O
E
=
1
6
Radius of incircle of
∆
P
Q
R
=
√
3
2
(
2
−
√
3
)
Area of
∆
S
O
E
=
√
3
12
Length of
R
S
=
√
7
2
Validate
Solution:
√
3
sinp
=
2
⇒
sinp
=
√
3
2
so
P
=
60
∘
,
120
∘
1
sinQ
=
2
⇒
sinQ
=
1
2
so
Q
=
30
∘
,
150
∘
Clearly P = 60º not possible. So P = 120º, Q = 30º, R = 30º
P
Q
=
P
R
=
1
,
Q
E
=
R
E
=
√
3
2
O must be centroid so
O
E
=
1
3
⋅
P
E
=
1
3
√
1
−
3
4
=
1
6
option 1 is correct
Area of
∆
P
Q
R
=
1
2
×
√
3
×
1
2
=
√
3
4
Semi-perimeter
=
(
√
3
+
1
+
1
)
2
=
√
3
+
2
2
So in radius of
∆
P
Q
R
=
√
3
4
√
3
+
2
2
=
√
3
(
2
−
√
3
)
2
option 2 is correct
R
S
=
√
P
R
2
+
Q
R
2
−
2
P
S
2
2
=
√
1
+
3
−
2
×
1
4
2
=
√
7
2
option 4 is correct
Area of
∆
S
O
E
=
1
3
.
Area of ΔSPE
=
1
3
×
1
4
5
area of ΔPQR
=
1
6
×
1
2
×
√
3
4
=
√
3
48
option 3 is incorrect
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