Properties of Triangles 3 Question 9
9. In a $\triangle P Q R$, let $\angle P Q R=30^{\circ}$ and the sides $P Q$ and $Q R$ have lengths $10 \sqrt{3}$ and 10 , respectively. Then, which of the following statement(s) is (are) TRUE? (2018 Adv)
(a) $\angle Q P R=45^{\circ}$
(b) The area of the $\triangle P Q R$ is $25 \sqrt{3}$ and $\angle Q R P=120^{\circ}$
(c) The radius of the incircle of the $\triangle P Q R$ is $10 \sqrt{3}-15$
(d) The area of the circumcircle of the $\triangle P Q R$ is $100 \pi$
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Answer:
Correct Answer: 9. $(b, c, d)$
Solution:
- We have, In $\triangle PQR$
$$ \begin{aligned} \angle P Q R & =30^{\circ} \\ P Q & =10 \sqrt{3} \\ Q R & =10 \end{aligned} $$
By cosine rule
$$ \begin{array}{rlrl} & \cos 30^{\circ} & =\frac{PQ^{2}+QR^{2}-PR^{2}}{2 PQ \cdot QR} \\ \Rightarrow \quad & \frac{\sqrt{3}}{2} & =\frac{300+100-PR^{2}}{200 \sqrt{3}} \\ \Rightarrow \quad 300 & =300+100-PR^{2} \\ \Rightarrow \quad PR & =10 \\ \text { Since, PR }=QR=10 \\ \therefore \quad & \angle QPR & =30^{\circ} \text { and } \angle QRP=120^{\circ} \\ & & \text { Area of } \triangle PQR & =\frac{1}{2} PQ \cdot QR \cdot \sin 30^{\circ} \\ & =\frac{1}{2} \times 10 \sqrt{3} \times 10 \times \frac{1}{2}=25 \sqrt{3} \end{array} $$
Radius of incircle of
$$ \Delta PQR=\frac{\text { Area of } \triangle PQR}{\text { Semi - perimetre of } \Delta PQR} $$
$$ \begin{array}{cc} \text { i.e. } & r=\frac{\Delta}{s}=\frac{25 \sqrt{3}}{\frac{10 \sqrt{3}+10+10}{2}}=\frac{25 \sqrt{3}}{5(\sqrt{3}+2)} \\ \Rightarrow \quad r=5 \sqrt{3}(2-\sqrt{3}) \\ =10 \sqrt{3}-15 \end{array} $$
and radius of circumcircle $(R)=\frac{a b c}{4 \Delta}=\frac{10 \sqrt{3} \times 10 \times 10}{4 \times 25 \sqrt{3}}=10$
$\therefore$ Area of circumcircle of
$$ \Delta PQR=\pi R^{2}=100 \pi $$
Hence, option (b), (c) and (d) are correct answer.
