In the above figure, O is the center of the circle and, M and N lie on the circle. The area of the right triangle MON is $50 \text{ cm}^2$. 
What is the area of the circle in $\text{cm}^2$ ?
To find the area of the circle, we need to first understand the relationship between the right triangle \( \triangle MON \) and the circle.
Given:
Since \( \triangle MON \) is a right triangle with hypotenuse \( MN \) lying on the circle, \( O \) must be the midpoint of the hypotenuse \( MN \). According to the properties of the circle, this implies that the hypotenuse \( MN \) is the diameter of the circle.
The formula for the area of a right triangle is:
Area = \frac{1}{2} \times \text{base} \times \text{height}
In \( \triangle MON \), the base \( MO \) and height \( ON \) are both radii of the circle. Let's denote the radius as \( r \).
Thus, we have:
Area = \frac{1}{2} \times r \times r = 50
\frac{1}{2} r^2 = 50
r^2 = 100
Now, the area of the circle is given by the formula:
\text{Area of Circle} = \pi r^2
Substituting \( r^2 = 100 \):
\text{Area of Circle} = \pi \times 100 = 100\pi \, \text{cm}^2
Therefore, the area of the circle is \( 100\pi \, \text{cm}^2 \), which matches the option \( 100\pi \).
In the given figure, $P, Q$, and $R$ are three points on a circle of radius 10 cm with $O$ as its center, $\overline{PQ} = \overline{RQ}$, and $\angle PQR = 45^\circ$. The figure is representative.
The area of the shaded region $PQRO$ is ______________ cm$^2$.
