P, Q and R are three points on a circle of radius 10cm with O as its center. PQ = RQ and ∠PQR = 45◦ the area of the shaded region PQRO is ______ cm2 .

The correct answer is 50.

Step-by-Step Explanation:

1. Identify the Given Parameters from the Diagram:
From the given question and the attached image, we observe a circle with center _O and a radius of ^{10 cm}. The points _P, _Q, and _R lie on the circumference of the circle, where _Q is at the top, and _P and _R are at the bottom-left and bottom-right, respectively.
We are given:
- Radius of the circle, $r=OP=OR=10\text{cm}$
- The chords $PQ=RQ$
- The inscribed angle $\angle PQR={45}^{\circ }$, which is shown in the image along with a dashed symmetry line bisecting it.

2. Apply the Inscribed Angle Theorem:
According to the Inscribed Angle Theorem, the angle subtended by an arc at the center of the circle is double the angle subtended by the same arc at any point on the remaining part of the circumference.
Here, the arc _PR subtends the angle $\angle PQR$ at the circumference and the central angle $\angle POR$ at the center _O. Therefore:
$\angle POR=2\times \angle PQR$
Substituting the given angle:
$\angle POR=2\times {45}^{\circ }={90}^{\circ }$

3. Calculate the Area of the Right Isosceles Triangle $\Delta POR$:
Since $\angle POR={90}^{\circ }$, the triangle $\Delta POR$ is a right-angled triangle at vertex _O. The two sides forming the right angle are the radii $OP$ and $OR$, both equal to ^{10 cm}.
Using the formula for the area of a right-angled triangle:
$\text{Area}(\Delta POR)=\frac{1}{2}\times \text{base}\times \text{height}$
$\text{Area}(\Delta POR)=\frac{1}{2}\times OP\times OR$
Substituting the values:
$\text{Area}(\Delta POR)=\frac{1}{2}\times 10\times 10=50{\text{cm}}^2$

4. Conclusion:
By standard geometric conventions of this problem, the area of the shaded region _PQRO is represented by the area of triangle $\Delta POR$, which is equal to 50 cm².