Question

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

• A. $\angle QPR={45}^{\circ }$
• B. The area of the triangle $PQR$ is $25\sqrt{3}$ and $\angle QRP={120}^{\circ }$
• C. The radius of the incircle of the triangle $PQR$ is $10\sqrt{3}-15$
• D. The area of the circumcircle of the triangle $PQR$ is $100\pi$

Answer 1

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

The correct options are
B The area of the triangle $PQR$ is $25\sqrt{3}$ and $\angle QRP={120}^{\circ }$
C The radius of the incircle of the triangle $PQR$ is $10\sqrt{3}-15$
D The area of the circumcircle of the triangle $PQR$ is $100\pi$


By cosine law,
$\cos {30}^{\circ }=\frac{100+300-(PR{)}^2}{2\times 10\times 10\sqrt{3}}$
$\Rightarrow PR=10$
$\Rightarrow PR=QR,\text{then}\angle QPR=\angle PQR$
$\therefore \angle QPR={30}^{\circ }\text{and}\angle QRP={120}^{\circ }$
$\text{area}\left(PQR\right),△=\frac{1}{2}\times 10\sqrt{3}\times 10\sin {30}^{\circ }=25\sqrt{3}$

Radius of incircle is given by
$r=\frac{△}{S}=\frac{25\sqrt{3}}{\frac{1}{2}(10+10+10\sqrt{3})}=10\sqrt{3}-15$

Radius of the circumcircle is
$R=\frac{PR}{2\sin {30}^{\circ }}=10$
Therefore, area of the circumcircle $=\pi (10{)}^2=100\pi$