Question

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 ______________ \(\text{cm}^2\).

• A. 50
• B. \(25\sqrt{2}\)
• C. \(50\sqrt{2}\)
• D. 100

Hint:

Split the quadrilateral along \(OQ\) into two congruent triangles, each with two 10 cm sides and a 135 degree included angle.

Answer 1

The Correct Option is C

Step 1: Understand the shape of the shaded region.
The shaded region \(PQRO\) is a quadrilateral traced by going P to Q, Q to R, R to O, then O back to P. Since \(O\) is the center, \(OP = OQ = OR = 10\) cm, each being a radius. Drawing the diagonal \(OQ\) splits this quadrilateral into two triangles, \(\triangle OPQ\) and \(\triangle OQR\).

Step 2: Show the diagonal \(OQ\) bisects the given angle.
We are told \(PQ = RQ\), and also \(OP = OR\) (both radii). So triangles \(OPQ\) and \(ORQ\) share the side \(OQ\) and have \(OP=OR\), \(QP=QR\), which by the SSS rule makes them congruent. This congruence means \(OQ\) splits the 45 degree angle \(\angle PQR\) into two equal halves, so \(\angle OQP = \angle OQR = 22.5^{\circ}\).

Step 3: Find the angle at the center for one triangle.
In \(\triangle OPQ\), the sides \(OP\) and \(OQ\) are equal (both 10 cm radii), so this triangle is isosceles, and the angles opposite these equal sides are equal too. That means \(\angle OPQ = \angle OQP = 22.5^{\circ}\). Since the three angles of a triangle add to 180 degrees, \(\angle POQ = 180^{\circ} - 22.5^{\circ} - 22.5^{\circ} = 135^{\circ}\).

Step 4: Compute the area of \(\triangle OPQ\).
For a triangle with two known sides and the included angle, area \(= \frac{1}{2} \times \text{side}_1 \times \text{side}_2 \times \sin(\text{included angle})\). Here both sides are 10 cm and the included angle at \(O\) is 135 degrees:
\[ \text{Area}(\triangle OPQ) = \frac{1}{2}(10)(10)\sin(135^{\circ}) = 50 \times \frac{\sqrt{2}}{2} = 25\sqrt{2} \text{ cm}^2 \]

Step 5: Add the second triangle by symmetry.
Because \(\triangle OPQ\) and \(\triangle OQR\) are congruent (shown in Step 2), \(\triangle OQR\) also has area \(25\sqrt{2}\) cm\(^2\). Adding both pieces gives the full shaded quadrilateral:
\[ \text{Area}(PQRO) = 25\sqrt{2} + 25\sqrt{2} = 50\sqrt{2} \text{ cm}^2 \]

Final Answer:
The area of the shaded region is \(50\sqrt{2}\) cm\(^2\), option (C). \[ \boxed{50\sqrt{2} \text{ cm}^2} \]