In a circular museum hall of radius 14 m, some statues are displayed. Statues are kept inside the inner concentric circle of radius 7 m. One such statue lying in sector OAB, is fenced along line segments OA, AP, PB and BO where P is a point on outer circle. Based on above information, answer the following questions: 37(i)Find \(m\angle AOP\).
Show Hint
Whenever the adjacent side is exactly half the length of the hypotenuse in a right-angled triangle, the angle between them is always \(60^\circ\) because \(\cos 60^\circ = \frac{1}{2}\).
This is a standard property of \(30^\circ-60^\circ-90^\circ\) triangles.
Step 1: Understand the setup. Outer circle radius $R = 14\text{ m}$, inner circle radius $r = 7\text{ m}$. OA is a radius of the inner circle ($OA = 7\text{ m}$) and OP is a radius of the outer circle ($OP = 14\text{ m}$). The sector OAB is part of the inner circle and P lies on the outer circle. The fenced region follows OA, AP, PB, and BO. Step 2: Identify the sector angle. Since $OA = OB = 7\text{ m}$ and the sector is OAB, assume $\angle AOB = 60^\circ$ (giving equilateral triangle OAB where OA = OB = AB = 7 m). Step 3: Calculate the area of the sector OPB of the outer circle. \[ A_{\text{sector OPB}} = \frac{60}{360} \times \pi R^2 = \frac{1}{6} \times \frac{22}{7} \times 196 = \frac{1}{6} \times \frac{22 \times 196}{7} = \frac{22 \times 28}{6} = \frac{616}{6} \approx 102.67\text{ m}^2 \] Step 4: Calculate the area of the sector OAB of the inner circle. \[ A_{\text{sector OAB}} = \frac{60}{360} \times \pi r^2 = \frac{1}{6} \times \frac{22}{7} \times 49 = \frac{1}{6} \times 22 \times 7 = \frac{154}{6} \approx 25.67\text{ m}^2 \] Step 5: Find the area of the fenced region (annular sector). \[ A_{\text{fenced}} = A_{\text{sector OPB}} - A_{\text{sector OAB}} = 102.67 - 25.67 = 77\text{ m}^2 \] Step 6: Conclusion. The area of the fenced region is 77 m$^2$. \[ \boxed{\text{Area of fenced region} = 77 \text{ m}^2} \]
