To solve for the area of $\triangle QDR$, let's analyze the given information:

1. Circle Properties and Tangent:

   • $O$ is the center of the circle, and $QT$ is a tangent to the circle at point $Q$.
   • Since $QT$ is tangent at $Q$, $\angle OQT = 90^\circ$.

2. Given Information:

   • $\angle AQT = 45^\circ$, so $\angle OQR = 45^\circ$ because $AQT$ and $OQR$ share the same arc in the circle.
   • $AQ = 20 \, \text{cm}$.

3. Right Triangle $\triangle OQA$:

   • Since $OQ$ is perpendicular to the tangent $QT$ at point $Q$, $\triangle OQA$ is a right triangle with $\angle QOA = 45^\circ$.
   • Since $\triangle OQA$ is a right triangle and $\angle QOA = 45^\circ$, $\angle OAQ = 45^\circ$ as well, making $\triangle OQA$ an isosceles right triangle.
   • Thus, $OQ = AQ / \sqrt{2} = 20 / \sqrt{2} = 10\sqrt{2}$ cm.

4. Finding Area of $\triangle QDR$:

   • Since $OQ = OR$, $\triangle QDR$ is an isosceles right triangle (with $\angle OQR = 45^\circ$).
   • The area of $\triangle QDR$ can be calculated using the base $OQ = 10\sqrt{2}$ cm and height $QR = 10\sqrt{2}$ cm.

   $\text{Area of } \triangle QDR = \frac{1}{2} \times OQ \times QR = \frac{1}{2} \times 10\sqrt{2} \times 10\sqrt{2} = \frac{1}{2} \times 200 = 100 \, \text{cm}^2$

Final Answer:

The area of $\triangle QDR$ is $100 \, \text{cm}^2$.

Would you like further details or have any questions about this solution?

Here are some related questions to expand on this concept:

1. What is the significance of the tangent property in circle geometry?
2. How can you determine the length of other segments in a similar circle problem with tangents?
3. How does the concept of isosceles right triangles apply to circle geometry?
4. What other properties of circles and tangents could be useful in solving related problems?
5. How would the solution change if $\angle AQT$ was different?

Tip: In geometry problems involving circles, identifying tangents and radii relationships is key, as tangents are always perpendicular to the radius at the point of contact.