Question

# In a non-right-angled triangle △ 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, which of the following options is/are correct?Length of RS = √72Area of △ SOE=√312Length of OE = 16Radius of incircle of △ PQR=√32(2−√3)

Solution

## The correct options are A Length of RS = √72 C Length of OE = 16 D Radius of incircle of △ PQR=√32(2−√3) Using Sine law, ⇒PRsinQ=QRsinP=2R=2 PR=q=1,  p=QR=√3 sinP=√32⇒∠P=π3 or 2π3 sinQ=12⇒∠Q=π6 or 5π6 Since p>q⇒∠P>∠Q Hence, ∠P=2π3 and ∠Q=∠R=π/6 PQ=2RsinR=2.1.sinπ6=1 Using Apollonius' theorem 2((RS)2+(PS)2)=(PR)2+(RQ)2 2((RS)2+(12)2)=(1)2+(√3)2 Therefore, Length of RS=√72 Radius of incircle r=4RsinP2.sinQ2.sinR2=4.1.√32sin215o r=√32(2−√3) From Fig. we can observe that, (PE)2+(RE)2=(PR)2 ⇒(PE)2=1−34 PE=12 From △PQR, using property  OEPE=13⇒OE=16 Now, area of △SEF=14× area of △PQR =14.12.√3.12=√316 area of △SOE=13× area of △SEF=112× area of △PQR =112.12.√3.12=√348

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