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Mathematics · properties of traingles

JEE Advanced 2018 — Paper 1 — Question 28

In a triangle PQR , let ∠PQR=30∘\angle \mathrm{PQR}=30^{\circ} and the sides PQ and QR have lengths 10310 \sqrt{3} and 10 , respectively. Then, which of the following statement(s) is (are) TRUE ?

  1. Option A:

    ∠QPR=45∘\angle \mathrm{QPR}=45^{\circ}

  2. Option B:

    The area of the triangle PQR is 25325 \sqrt{3} and ∠QRP=120∘\angle \mathrm{QRP}=120^{\circ}

    Correct
  3. Option C:

    The radius of the incircle of the triangle PQR is 103−1510 \sqrt{3}-15

    Correct
  4. Option D:

    The area of the circumcircle of the triangle PQR is 100π100 \pi

    Correct

Answer: B, C, D

Step-by-step solution

(A) 10sin⁡θ=103sin⁡(150−θ)\frac{10}{\sin \theta}=\frac{10 \sqrt{3}}{\sin (150-\theta)}

⇒cos⁡θ2+32sin⁡θ=3sin⁡θ\Rightarrow \frac{\cos \theta}{2}+\frac{\sqrt{3}}{2} \sin \theta=\sqrt{3} \sin \theta

⇒13=tan⁡θ⇒θ=30∘\Rightarrow \frac{1}{\sqrt{3}}=\tan \theta \Rightarrow \theta=30^{\circ}

(B) Δ=1210.103sin⁡30∘=253,∠QRP=120∘\Delta=\frac{1}{2} 10.10 \sqrt{3} \sin 30^{\circ}=25 \sqrt{3}, \angle \mathrm{QRP}=120^{\circ}

(C) r=Δs=25310+53=253(10−53)25=103−15\mathrm{r}=\frac{\Delta}{\mathrm{s}}=\frac{25 \sqrt{3}}{10+5 \sqrt{3}}=\frac{25 \sqrt{3}(10-5 \sqrt{3})}{25}=10 \sqrt{3}-15

(D) Area =πR2=π4(101/2)2=100π=\pi R^{2}=\frac{\pi}{4}\left(\frac{10}{1 / 2}\right)^{2}=100 \pi

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 1
Subject
Mathematics
Chapter
properties of traingles
Topic
Incircles, excircles and its related properties