Step-by-step Solution

1. Interpret the Given Information

The question provides information about a triangle PQR. Let's use standard notation where sides opposite to vertices P, Q, R are denoted by p, q, r respectively.

• The phrase "In a $\Delta$ PQR = 30 $^\circ$" is poorly formatted but standard interpretation in this context is that one of the angles is 30$^{\circ}$. The most likely intended angle is $\angle PQR$, the angle at vertex Q. So, we assume $\angle Q = 30^\circ$.
• Side PQ is the side connecting vertices P and Q. In standard notation, this is side r. So, $r = 10\sqrt{3}$.
• Side QR is the side connecting vertices Q and R. In standard notation, this is side p. So, $p = 10$.

So we have a triangle with two sides and the included angle (SAS case): $p=10$, $r=10\sqrt{3}$, and $\angle Q = 30^\circ$. This defines a unique triangle.

2. Calculate the Third Side (PR = q)

We can use the Law of Cosines to find the length of the side q (PR): $q^2 = p^2 + r^2 - 2pr \cos Q$ Substituting the given values: $q^2 = (10)^2 + (10\sqrt{3})^2 - 2(10)(10\sqrt{3}) \cos 30^\circ$ $q^2 = 100 + 100(3) - 200\sqrt{3} \left(\frac{\sqrt{3}}{2}\right)$ $q^2 = 100 + 300 - 100(3)$ $q^2 = 400 - 300 = 100$ $q = \sqrt{100} = 10$ So, the length of side PR is 10.

3. Calculate the Remaining Angles ($\\/angle P$ and $\\/angle R$)

The sides of the triangle are $p=10$, $q=10$, and $r=10\sqrt{3}$. Since sides $p$ and $q$ are equal ($p=q=10$), the triangle is isosceles. The angles opposite to these equal sides must also be equal.

• Angle opposite to side p (QR) is $\angle P$ (or $\angle QPR$).
• Angle opposite to side q (PR) is $\angle Q$ (or $\angle PQR$). Therefore, $\angle P = \angle Q$. Since we are given $\angle Q = 30^\circ$, it follows that $\angle P = 30^\circ$.

The sum of angles in a triangle is $180^\circ$. So, we can find $\angle R$ (or $\angle QRP$): $\angle R = 180^\circ - (\angle P + \angle Q)$ $\angle R = 180^\circ - (30^\circ + 30^\circ) = 180^\circ - 60^\circ = 120^\circ$

Summary of triangle properties:

• Sides: $p=10$, $q=10$, $r=10\sqrt{3}$
• Angles: $\angle P = 30^\circ$, $\angle Q = 30^\circ$, $\angle R = 120^\circ$

4. Evaluate Each Option

A: $\angle QPR = 45^\circ$ $\angle QPR$ is the angle at vertex P. We found $\angle P = 30^\circ$. Since $30^\circ \neq 45^\circ$, this statement is FALSE.

B: The area of the $\Delta PQR$ is $25\sqrt 3$ and $\angle QRP = 120^\circ$ First part: $\angle QRP$ is the angle at vertex R. We found $\angle R = 120^\circ$. This is correct. Second part: The area of the triangle can be calculated using the formula Area = $\frac{1}{2}pr \sin Q$. $\text{Area} = \frac{1}{2}(10)(10\sqrt{3}) \sin 30^\circ = \frac{1}{2}(100\sqrt{3})\left(\frac{1}{2}\right) = 25\sqrt{3}$This is also correct. Since both parts of the statement are true, option B is TRUE.

C: The radius of the incircle of the $\Delta PQR$ is $10\sqrt 3 - 15$ The radius of the incircle ($r_{in}$) is given by the formula $r_{in} = \frac{\text{Area}}{s}$, where $s$ is the semi-perimeter. First, calculate the semi-perimeter $s$: $s = \frac{p+q+r}{2} = \frac{10+10+10\sqrt{3}}{2} = \frac{20+10\sqrt{3}}{2} = 10+5\sqrt{3}$ Now, calculate the inradius: $r_{in} = \frac{25\sqrt{3}}{10+5\sqrt{3}} = \frac{25\sqrt{3}}{5(2+\sqrt{3})} = \frac{5\sqrt{3}}{2+\sqrt{3}}$ To simplify, rationalize the denominator: $r_{in} = \frac{5\sqrt{3}(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} = \frac{10\sqrt{3} - 5(3)}{2^2 - (\sqrt{3})^2} = \frac{10\sqrt{3}-15}{4-3} = 10\sqrt{3}-15$ This statement is TRUE.

D: The area of the circumcircle of the $\Delta PQR$ is 100 $\pi$ The radius of the circumcircle ($R_{circ}$) can be found using the Law of Sines: $2R_{circ} = \frac{a}{\sin A}$. Using side q and angle Q: $2R_{circ} = \frac{q}{\sin Q} = \frac{10}{\sin 30^\circ} = \frac{10}{1/2} = 20$ $R_{circ} = 10$ The area of the circumcircle is $\pi R_{circ}^2$. $\text{Area} = \pi (10)^2 = 100\pi$ This statement is TRUE.

Conclusion Based on the analysis, statements B, C, and D are true.