Akshay places three identical rings of diameter $21 ext{ cm}$ in such a manner that each one touches the other two rings. He observes that there is a space enclosed by the circles. What is the area (in $ ext{cm}^2$) of this portion? (Take $\pi = \frac{22}{7}$ and $\sqrt{3} = 1.732$)

**Note:** Round off the answer to the nearest 2 decimal places.

Question

Akshay places three identical rings of diameter $21 	ext{ cm}$ in such a manner that each one touches the other two rings. He observes that there is a space enclosed by the circles. What is the area (in $	ext{cm}^2$) of this portion? (Take $\pi = \frac{22}{7}$ and $\sqrt{3} = 1.732$)

**Note:** Round off the answer to the nearest 2 decimal places.

Accepted Answer

**Solution:**

1. Given diameter of each ring = $21 	ext{ cm}$, so radius $r = \frac{21}{2} = 10.5 	ext{ cm}$.

2. The centers of the three circles form an equilateral triangle with side length equal to the diameter of the circles, i.e., side $= 21 	ext{ cm}$.

3. Calculate the area of the equilateral triangle:
   $$	ext{Area of triangle} = \frac{\sqrt{3}}{4} 	imes (	ext{side})^2 = \frac{1.732}{4} 	imes 21^2 = \frac{1.732}{4} 	imes 441 = 190.953 	ext{ cm}^2$$

4. Each sector formed inside the triangle by the circle is $60^\circ$ (since the triangle is equilateral, each angle is $60^\circ$).

5. Area of one sector:
   $$= \frac{60}{360} 	imes \pi r^2 = \frac{1}{6} 	imes \frac{22}{7} 	imes 10.5^2 = \frac{1}{6} 	imes \frac{22}{7} 	imes 110.25 = 57.75 	ext{ cm}^2$$

6. Area of three such sectors:
   $$3 	imes 57.75 = 173.25 	ext{ cm}^2$$

7. Area of the enclosed portion (area inside the triangle but outside the sectors):
   $$= 	ext{Area of triangle} - 	ext{Area of 3 sectors} = 190.953 - 173.25 = 17.703 \approx 17.70 	ext{ cm}^2$$

**Answer:** $17.70 	ext{ cm}^2$

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