Question

In a $△PQR$ let $\angle$PQR$={30}^o$ 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 $={45}^o$
• B. The area of the triangle PQR is $25\sqrt{3}$ and $\angle$QRP$={120}^o$
• 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$={120}^o$
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 $100\pi$

(A) $\cos {30}^o=\frac{(10\sqrt{3}{)}^2+(10{)}^2-(PR{)}^2}{2\times 10\sqrt{3}\times 10}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{400-(PR{)}^2}{200\sqrt{3}}$

$\Rightarrow PR=10$
$\because QR=PR$ $\Rightarrow PQR=\angle QPR$
$\angle$QPR $={30}^o$
(B) Area of $△$PQR $=\frac{1}{2}\times 10\sqrt{3}\times 10\times \sin {30}^o=\frac{1}{2}\times 10\times 10\sqrt{3}\times \frac{1}{2}$
$=25\sqrt{3}$
$\angle QRP={180}^o-({30}^o+{30}^o)={120}^o$
(C) $r=\frac{△}{S}=\frac{25\sqrt{3}}{\left(\frac{10+10+10\sqrt{3}}{2}\right)}=\frac{25\sqrt{3}}{10+5\sqrt{3}}$
$=5\sqrt{3}\cdot (2-\sqrt{3})=10\sqrt{3}-15$
(D) $R=\frac{a}{2\sin A}=\frac{10}{2\sin {30}^o}=10$
$\therefore$ Area $=\pi R^2=100\pi$