Find the area of the shaded region in the given figure, where a circular arc of radius $6\text{ cm}$ has been drawn with vertex $O$ of an equilateral triangle $OAB$ of side $12\text{ cm}$ as centre.
Step-by-Step Solution
Key Concept: Total area $= \text{Area of major sector } (300^\circ) + \text{Area of equilateral } \Delta = \dfrac{300}{360} \times \dfrac{22}{7} \times 36 + \dfrac{\sqrt{3}}{4} \times 144 = \dfrac{5}{6} \times \dfrac{792}{7} + 36\sqrt{3} = \dfrac{660}{7} + 36\sqrt{3} = 94.28 + 62.35 = 156.63\text{ cm}^2$.
Major sector angle $= 360^\circ - 60^\circ = 300^\circ$. $\text{Major Sector Area} = \dfrac{5}{6} \times \dfrac{22}{7} \times 36 = \dfrac{660}{7} \approx 94.28\text{ cm}^2$. [1.5 Marks]
$\text{Triangle Area} = \dfrac{\sqrt{3}}{4} \times 144 = 36\sqrt{3} \approx 62.35\text{ cm}^2$. [1.0 Mark]
Total shaded area $= \dfrac{660}{7} + 36\sqrt{3} \approx 156.63\text{ cm}^2$. [0.5 Mark]
---
🎯 Official CBSE Marking Scheme:
Calculating major sector area $= 660/7\text{ cm}^2$: 1.5 Marks
Calculating triangle area $= 36\sqrt{3}\text{ cm}^2$: 1.0 Mark
Summing to get $156.63\text{ cm}^2$: 0.5 Mark
Correct Answer: