The Olympic symbol comprising five interlocking rings represents the union of the five continents of the world and the meeting of athletes from all over the world at the Olympic games. In order to spread awareness about Olympic games, students of Class-X took part in various activities organised by the school. One such group of students made 5 circular rings in the school lawn with the help of ropes. Each circular ring required 44 m of rope. Also, in the shaded regions as shown in the figure, students made rangoli showcasing various sports and games. It is given that $\triangle OAB$ is an equilateral triangle and all unshaded regions are congruent.
Generated by claude-sonnet-4-6 · 2026-06-15 10:38 · grounding stimulus
Model Answer
(i) Radius of each circular ring:
Circumference = 44 m
$$2\pi r = 44 \implies r = \frac{44}{2 \times \frac{22}{7}} = \frac{44 \times 7}{44} = 7 \text{ m}$$
Radius = 7 m
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(ii) Measure of ∠AOB:
Since △OAB is equilateral and O is the centre with OA = OB = r (radii), and AB = r (side of equilateral triangle),
$$\angle AOB = 60°$$
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(iii) Area of shaded region R₁:
R₁ = Area of sector OAB − Area of △OAB
$$\text{Area of sector} = \frac{60°}{360°} \times \pi r^2 = \frac{1}{6} \times \frac{22}{7} \times 49 = \frac{77}{3} \text{ m}^2$$
$$\text{Area of } \triangle OAB = \frac{\sqrt{3}}{4} \times r^2 = \frac{\sqrt{3}}{4} \times 49 = \frac{49\sqrt{3}}{4} \text{ m}^2$$
$$R_1 = \frac{77}{3} - \frac{49\sqrt{3}}{4} = \left(\frac{77}{3} - \frac{49\sqrt{3}}{4}\right) \text{ m}^2$$
Source: Areas Related to Circles, Case Study
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Explanation
- For (i), use circumference formula directly — examiner expects one clean step.
- For (ii), since OA = OB = AB = r (all equal to radius, forming an equilateral triangle), ∠AOB = 60°. State this reasoning briefly.
- For (iii), the standard approach is sector area − triangle area. Use θ = 60°, r = 7. Leave answer in exact form (with √3); do not approximate unless asked. Both the setup and the subtraction fetch marks.