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 \(\text{cm}^2\).
Show Hint
Use the inscribed angle theorem to get \(\angle POR=90^{\circ}\), split the rest of the circle equally using \(PQ=RQ\), then add the areas of triangles \(OPQ\) and \(OQR\) using \(\frac{1}{2}ab\sin\theta\).
Step 1: Place the circle on coordinate axes.
Put the center $O$ at the origin, with the circle having radius $10$. Since $\angle PQR = 45^{\circ}$ is an inscribed angle standing on arc $PR$ (the arc not containing $Q$), the central angle is $\angle POR = 2 \times 45^{\circ} = 90^{\circ}$ by the inscribed angle theorem.
Step 2: Use the equal chords to fix each point's angle around the circle.
Because $PQ = RQ$, point $Q$ sits symmetrically between $P$ and $R$, splitting the remaining $360^{\circ}-90^{\circ}=270^{\circ}$ of arc equally: $135^{\circ}$ from $Q$ to $P$ and $135^{\circ}$ from $Q$ to $R$. Put $Q$ straight up at angle $90^{\circ}$:
$$Q = (10\cos 90^{\circ}, 10\sin 90^{\circ}) = (0,10)$$
$P$ is $135^{\circ}$ further around (angle $225^{\circ}$) and $R$ is $135^{\circ}$ the other way (angle $-45^{\circ}$):
$$P = (10\cos 225^{\circ}, 10\sin 225^{\circ}) = (-5\sqrt2,-5\sqrt2)$$
$$R = (10\cos(-45^{\circ}), 10\sin(-45^{\circ})) = (5\sqrt2,-5\sqrt2)$$
Step 3: Apply the shoelace formula to quadrilateral $P\text{-}Q\text{-}R\text{-}O$.
For vertices taken in order, the shoelace formula gives $\text{Area} = \frac{1}{2}\left| \sum (x_i y_{i+1} - x_{i+1} y_i) \right|$. Using $P(-5\sqrt2,-5\sqrt2)$, $Q(0,10)$, $R(5\sqrt2,-5\sqrt2)$, $O(0,0)$:
$$x_P y_Q - x_Q y_P = (-5\sqrt2)(10)-(0)(-5\sqrt2) = -50\sqrt2$$
$$x_Q y_R - x_R y_Q = (0)(-5\sqrt2)-(5\sqrt2)(10) = -50\sqrt2$$
$$x_R y_O - x_O y_R = 0, \qquad x_O y_P - x_P y_O = 0$$
Step 4: Add the terms and take half the absolute value.
$$\text{Sum} = -50\sqrt2-50\sqrt2+0+0 = -100\sqrt2$$
$$\text{Area} = \frac{1}{2}|-100\sqrt2| = 50\sqrt2 \text{ cm}^2$$
Final Answer:
Placing the points on coordinate axes and using the shoelace formula gives the same result as the two-triangle method: the shaded area is $50\sqrt2$ cm$^2$.
\[ \boxed{50\sqrt2 \text{ cm}^2} \]
