To find the area of the shaded region \(PQRO\), we follow these steps:

1. Given that points \(P, Q,\) and \(R\) lie on a circle with radius 10 cm, and \( \overline{PQ} = \overline{RQ} \), \( \triangle PQR\) is isosceles.
2. The angle \( \angle PQR = 45^\circ \).
3. Since \(PQ = RQ\) and \(QO\) is a bisector, the angle \(\angle PQO = \angle RQO = \frac{45^\circ}{2} = 22.5^\circ\).
4. The line segments \(PQ\) and \(RQ\) are chords of the circle.
5. Using the property that the sum of angles in a triangle is \(180^\circ\), we have: \(\angle POQ = 180^\circ - 45^\circ = 135^\circ\\)
6. To find the area of \( \triangle POQ \) and \( \triangle ROQ \), use the formula for the area of a triangle: \(\text{Area} = \frac{1}{2} \times r^2 \times \sin(\theta)\) where \(\theta\) is the angle between the lines.
7. Calculate the area of each sector:
   • The area of sector \(POQ = \frac{1}{2} \times 10^2 \times \sin(135^\circ) = 50\sin(135^\circ)\).
   • Since \(\sin(135^\circ) = \sin(180^\circ - 135^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2}\), the area of sector \(POQ\) becomes \(50 \times \frac{\sqrt{2}}{2} = 25\sqrt{2}\) cm\(^2\).
   • The area of sector \(ROQ\) is equal to that of sector \(POQ\) due to symmetry: \(25\sqrt{2}\) cm\(^2\).
8. The total area of \( \triangle PQRO \) is the area of both sectors: \(\text{Area} = 25\sqrt{2} + 25\sqrt{2} = 50\sqrt{2} \text{ cm}^2\).

Thus, the area of the shaded region \(PQRO\) is \(50\sqrt{2} \text{ cm}^2\).