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: Identify what OP, OQ, OR represent.
\(P\), \(Q\), \(R\) lie on a circle of radius 10 cm with center \(O\). So \(OP=OQ=OR=10\) cm, since every radius of a circle has the same length.
Step 2: Use the inscribed angle theorem to find the central angle on the far arc.
\(\angle PQR=45^{\circ}\) is an inscribed angle at \(Q\), standing on the arc \(PR\) that does not contain \(Q\). The inscribed angle theorem says an inscribed angle is half the central angle standing on the same arc. So the central angle \(\angle POR\) (the one not passing through \(Q\)) is
\[ \angle POR = 2 \times 45^{\circ} = 90^{\circ} \]
Step 3: Use \(\overline{PQ}=\overline{RQ}\) to split the remaining arc equally.
Equal chords in the same circle cut off equal arcs. Since \(PQ=RQ\), arc \(PQ\) (not through \(R\)) equals arc \(QR\) (not through \(P\)). The rest of the circle, going from \(P\) to \(R\) the long way through \(Q\), covers \(360^{\circ}-90^{\circ}=270^{\circ}\). Splitting this equally between arc \(PQ\) and arc \(QR\):
\[ \angle POQ = \angle QOR = \frac{270^{\circ}}{2} = 135^{\circ} \]
Step 4: Break the shaded shape \(PQRO\) into two triangles.
The shaded region \(PQRO\) is a kite-shaped quadrilateral made of two triangles sharing the diagonal \(OQ\): triangle \(OPQ\) and triangle \(OQR\). In each triangle two sides are radii of length 10 cm, with included angle found above.
Step 5: Find the area of each triangle using the two-sides-and-included-angle formula.
For a triangle with sides \(a\), \(b\) and included angle \(\theta\), area \(=\frac{1}{2}ab\sin\theta\). Here \(a=b=10\) and \(\theta=135^{\circ}\) for both triangles:
\[ \text{Area}(OPQ) = \frac{1}{2}(10)(10)\sin 135^{\circ} = 50 \times \frac{\sqrt{2}}{2} = 25\sqrt{2} \text{ cm}^2 \]
By the same working, Area\((OQR) = 25\sqrt{2}\) cm\(^2\) too, since it has the same two side lengths and the same \(135^{\circ}\) angle.
Step 6: Add the two triangles to get the shaded area.
\[ \text{Area}(PQRO) = 25\sqrt{2}+25\sqrt{2} = 50\sqrt{2} \text{ cm}^2 \]
Final Answer:
The area of the shaded region \(PQRO\) is \(50\sqrt{2}\) cm\(^2\).
\[ \boxed{50\sqrt{2}\text{ cm}^2} \]
