Question

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 :

(i) Find m\(\angle\)AOP.

Hint:

Whenever the adjacent side of a right triangle is exactly half of the hypotenuse, the angle between them is always \(60^\circ\) because \(\cos 60^\circ = 0.5\).

Answer 1

Correct Answer: 60

Step 1: Understand the geometry.
Two concentric circles: inner radius \(r = 7\) m and outer radius \(R = 14\) m, both with centre O. Point P is on the outer circle. AP is tangent to the inner circle, with A and B on the inner circle. The statue is fenced along OA, AP, PB, BO.
Step 2: Find the length of the tangent AP.
Since AP is tangent to the inner circle (radius 7 m) from P on the outer circle (radius 14 m): \(AP = \sqrt{OP^2 - OA^2} = \sqrt{14^2 - 7^2} = \sqrt{196 - 49} = \sqrt{147} = 7\sqrt{3}\) m.
Step 3: Find angle AOP using trigonometry.
In triangle OAP (right-angled at A, since tangent is perpendicular to radius): \(\sin(\angle OPA) = \frac{OA}{OP} = \frac{7}{14} = \frac{1}{2}\), so \(\angle OPA = 30^\circ\). Therefore \(\angle AOP = 60^\circ\). By symmetry, \(\angle BOP = 60^\circ\), so total \(\angle AOB = 120^\circ\).
Step 4: Find the perimeter of the fenced region.
Perimeter \(= OA + AP + PB + BO = 7 + 7\sqrt{3} + 7\sqrt{3} + 7 = 14 + 14\sqrt{3} = 14(1 + \sqrt{3})\) m.
Step 5: Find the area of the fenced region.
Area \(=\) Area of sector OAB (angle \(120^\circ\)) \(+\) Area of \(\triangle OAP\) \(+\) Area of \(\triangle OBP\). Sector area \(= \frac{120}{360} \times \pi \times 7^2 = \frac{1}{3} \times \frac{22}{7} \times 49 = \frac{154}{3} \approx 51.33\) m\(^2\). Each triangle area \(= \frac{1}{2} \times 7 \times 7\sqrt{3} = \frac{49\sqrt{3}}{2}\) m\(^2\). Total area \(= \frac{154}{3} + 49\sqrt{3} \approx 51.33 + 84.87 = 136.2\) m\(^2\).
Step 6: State the final answers.
\[ \boxed{\text{Perimeter} = 14(1+\sqrt{3}) \text{ m} \approx 38.25 \text{ m}, \quad \text{Area} \approx 136.2 \text{ m}^2} \]