JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Properties of Triangle

In a non-right-angled triangle

Δ 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 = 3, q = 1

, and the radius of the circumcircle of the

△ PQR

equals 1 , then which of the following options is/are correct ?

(A)Area of $△ SOE = \frac{3}{12}$
(B)Radius of incircle of $Δ PQR = \frac{3}{2} (2 - 3)$ ✓ Correct
(C)Length of $RS = \frac{7}{2}$ ✓ Correct
(D)Length of $OE = \frac{1}{6}$ ✓ Correct

Answer: (B), (C), (D)

Solution

Using the sine rule:

\frac{s i n P}{3} = \frac{s i n Q}{1} = \frac{1}{2}

This gives

P = \frac{2 π}{3}

and

Q = R = \frac{π}{6}

(since

p > q

implies

P > Q

and the triangle is not right-angled).

Inradius:

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

✓

Length of median:

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

✓

Since O is the centroid,

OE = \frac{1}{3} PE = \frac{1}{3} \cdot \frac{3}{2} = \frac{1}{6}

✓

Area of

△ SOE = \frac{1}{12}

area of

△ PQR = \frac{1}{12} \cdot \frac{3}{4} = \frac{3}{48}

✗

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Properties of Triangle

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