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Mathematics · properties of traingles

JEE Advanced 2019 — Paper 1 — Question 34

In a non-right-angled triangle ΔPQR\Delta \mathrm{PQR}, let p,q,r\mathrm{p}, \mathrm{q}, \mathrm{r} denote the lengths of the sides opposite to the angles at P , Q,R\mathrm{Q}, \mathrm{R} respectively. The median from R meets the side PQ at S , the perpendicular from P meets the side QR at EE, and RS and PE intersect at OO. If p=3,q=1p=\sqrt{3}, q=1, and the radius of the circumcircle of the △PQR\triangle P Q R equals 1 , then which of the following options is/are correct?

  1. Option A:

    length of OE=16\mathrm{OE}=\frac{1}{6}

    Correct
  2. Option B:

    Radius of incircle of △PQR=32(2−3)\triangle \mathrm{PQR}=\frac{\sqrt{3}}{2}(2-\sqrt{3})

    Correct
  3. Option C:

    Length of RS=72\mathrm{RS}=\frac{\sqrt{7}}{2}

    Correct
  4. Option D:

    Are of △SOE=312\triangle \mathrm{SOE}=\frac{\sqrt{3}}{12}

Answer: A, B, C

Step-by-step solution

By sine rule

sin⁡PP=12R⇒sin⁡P=32⇒P=60∘ or 120∘sin⁡θ2=12R⇒sin⁡θ=12⇒θ=30∘,150∘\begin{aligned} & \frac{\sin \mathrm{P}}{\mathrm{P}}=\frac{1}{2 \mathrm{R}} \Rightarrow \sin \mathrm{P}=\frac{\sqrt{3}}{2} \Rightarrow \mathrm{P}=60^{\circ} \text { or } 120^{\circ} \\& \frac{\sin \theta}{2}=\frac{1}{2 \mathrm{R}} \Rightarrow \sin \theta=\frac{1}{2} \Rightarrow \theta=30^{\circ}, 150^{\circ} \end{aligned}

so only possible combination, P=120∘&Q=30∘⇒∠R=30∘\mathrm{P}=120^{\circ} \& \mathrm{Q}=30^{\circ} \Rightarrow \angle \mathrm{R}=30^{\circ}

(C) Length of RS ⇒(PR)2+(QR)2=2(RS2+PS2)⇒RS=72\Rightarrow(\mathrm{PR})^{2}+(\mathrm{QR})^{2}=2\left(\mathrm{RS}^{2}+\mathrm{PS}^{2}\right) \Rightarrow \mathrm{RS}=\frac{\sqrt{7}}{2}

(A) One △PQR=1/2PQ⋅PRsin⁡120∘=34⇒PE=1/2\triangle \mathrm{PQR}=1 / 2 \mathrm{PQ} \cdot \mathrm{PR} \sin 120^{\circ}=\frac{\sqrt{3}}{4} \Rightarrow \mathrm{PE}=1 / 2

⇒OE=1/3PE=1/6\Rightarrow \mathrm{OE}=1 / 3 \mathrm{PE}=1 / 6

(B) r=Δs=2Δp+q+r=2⋅343+2=32(2+3)=32(2−3)\mathrm{r}=\frac{\Delta}{\mathrm{s}}=\frac{2 \Delta}{\mathrm{p}+\mathrm{q}+\mathrm{r}}=\frac{2 \cdot \frac{\sqrt{3}}{4}}{\sqrt{3}+2}=\frac{\sqrt{3}}{2(2+\sqrt{3})}=\frac{\sqrt{3}}{2}(2-\sqrt{3})

(D) are of △OSE=1/2SE.OEsin⁡∠SEO\triangle \mathrm{OSE}=1 / 2 \mathrm{SE} . \mathrm{OE} \sin \angle \mathrm{SEO}

12⋅12⋅16sin⁡60∘=348 unit 2\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{6} \sin 60^{\circ}=\frac{\sqrt{3}}{48} \text { unit }^{2}
Solution figure

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Exam
JEE Advanced 2019
Paper
Paper 1
Subject
Mathematics
Chapter
properties of traingles
Topic
Incircles, excircles and its related properties