Geometric Properties and Incenter of Non Right Angled Triangle

To determine the correct options, we analyze the given triangle $\Delta PQR$.

Step 1: Determining the Angles and Sides of $\Delta PQR$

Using the Sine Rule: $\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R} = 2R_{\text{circ}}$ Given $p = \sqrt{3}$, $q = 1$, and $R_{\text{circ}} = 1$: $\sin P = \frac{p}{2R_{\text{circ}}} = \frac{\sqrt{3}}{2} \implies P = 60^\circ \text{ or } 120^\circ$ $\sin Q = \frac{q}{2R_{\text{circ}}} = \frac{1}{2} \implies Q = 30^\circ \text{ or } 150^\circ$

Since the sum of angles in a triangle is $180^\circ$:

1. If $P = 60^\circ$ and $Q = 30^\circ$, then $R = 90^\circ$, which contradicts the condition that $\Delta PQR$ is a non-right-angled triangle.
2. If $P = 60^\circ$ and $Q = 150^\circ$, $P + Q > 180^\circ$, which is impossible.
3. If $P = 120^\circ$ and $Q = 30^\circ$, then: $R = 180^\circ - (120^\circ + 30^\circ) = 30^\circ$ This gives a non-right-angled triangle with angles: $P = 120^\circ, \quad Q = 30^\circ, \quad R = 30^\circ$

Since $Q = R = 30^\circ$, the triangle is isosceles with: $r = q = 1, \quad p = \sqrt{3}$

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Step 2: Length of the Median $RS$ (Option A)

The segment $RS$ is the median drawn from vertex $R$ to the side $PQ$ of length $r = 1$. Using the median length formula: $RS = \frac{1}{2} \sqrt{2p^2 + 2q^2 - r^2}$ Substituting the values: $RS = \frac{1}{2} \sqrt{2(\sqrt{3})^2 + 2(1)^2 - 1^2} = \frac{1}{2} \sqrt{6 + 2 - 1} = \frac{\sqrt{7}}{2}$ Thus, Option A is correct.

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Step 3: Geometry of Point $O$ and Length of $OE$ (Option C)

Since $PQ = PR = 1$, $\Delta PQR$ is isosceles with vertex at $P$. The perpendicular $PE$ from $P$ to the base $QR$ is also the median to $QR$. Hence, $E$ is the midpoint of $QR$.

The altitude $PE$ has length: $PE = PR \sin(\angle R) = 1 \cdot \sin 30^\circ = \frac{1}{2}$

Since $RS$ is the median to $PQ$ and $PE$ is the median to $QR$, their point of intersection $O$ is the centroid of $\Delta PQR$.

The centroid divides each median in the ratio $2:1$ from the vertex. Therefore: $OE = \frac{1}{3} PE = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ Thus, Option C is correct.

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Step 4: Area of $\Delta SOE$ (Option B)

Let us set up a Cartesian coordinate system with $E$ at the origin $(0,0)$ and the base $QR$ along the $x$-axis:

• Since $E$ is the midpoint of $QR$ and $QR = \sqrt{3}$: $Q = \left(-\frac{\sqrt{3}}{2}, 0\right), \quad R = \left(\frac{\sqrt{3}}{2}, 0\right)$
• The vertex $P$ lies on the $y$-axis: $P = \left(0, \frac{1}{2}\right)$
• $S$ is the midpoint of $PQ$: $S = \left(\frac{0 - \frac{\sqrt{3}}{2}}{2}, \frac{\frac{1}{2} + 0}{2}\right) = \left(-\frac{\sqrt{3}}{4}, \frac{1}{4}\right)$
• Centroid $O$ lies on the segment $PE$ (the $y$-axis): $O = \left(0, \frac{1}{6}\right)$

The area of $\Delta SOE$ with vertices $E(0,0)$, $O\left(0, \frac{1}{6}\right)$, and $S\left(-\frac{\sqrt{3}}{4}, \frac{1}{4}\right)$ is: $\text{Area}(\Delta SOE) = \frac{1}{2} \left| x_E(y_O - y_S) + x_O(y_S - y_E) + x_S(y_E - y_O) \right|$ $\text{Area}(\Delta SOE) = \frac{1}{2} \left| 0 + 0 + \left(-\frac{\sqrt{3}}{4}\right)\left(0 - \frac{1}{6}\right) \right| = \frac{1}{2} \times \frac{\sqrt{3}}{24} = \frac{\sqrt{3}}{48}$ Since $\frac{\sqrt{3}}{48} \neq \frac{\sqrt{3}}{12}$, Option B is incorrect.

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Step 5: Radius of Incircle of $\Delta PQR$ (Option D)

The inradius $r_{\text{in}}$ is given by: $r_{\text{in}} = \frac{\Delta_{\text{area}}}{s}$

The total area of $\Delta PQR$ is: $\Delta_{\text{area}} = \frac{1}{2} qr \sin P = \frac{1}{2} \times 1 \times 1 \times \sin 120^\circ = \frac{\sqrt{3}}{4}$

The semi-perimeter $s$ is: $s = \frac{p + q + r}{2} = \frac{\sqrt{3} + 1 + 1}{2} = \frac{2 + \sqrt{3}}{2}$

Thus: $r_{\text{in}} = \frac{\frac{\sqrt{3}}{4}}{\frac{2 + \sqrt{3}}{2}} = \frac{\sqrt{3}}{2(2 + \sqrt{3})} = \frac{\sqrt{3}(2 - \sqrt{3})}{2(4 - 3)} = \frac{\sqrt{3}}{2}(2 - \sqrt{3})$ Thus, Option D is correct.

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Conclusion

The correct options are A, C, and D.