Akshay places three identical rings of diameter $21 ext{ cm}$ such that each ring touches the other two. Find the area (in $ ext{cm}^2$) of the space enclosed by the three rings. (Take $\pi = \frac{22}{7}$ and $\sqrt{3} = 1.732$)

**Note:** Round off your answer to 2 decimal places.

Question

Akshay places three identical rings of diameter $21 	ext{ cm}$ such that each ring touches the other two. Find the area (in $	ext{cm}^2$) of the space enclosed by the three rings. (Take $\pi = \frac{22}{7}$ and $\sqrt{3} = 1.732$)

**Note:** Round off your answer to 2 decimal places.

Accepted Answer

**Solution:**

![Screenshot 2025-08-31 at 10.48.05 PM.png](https://res.cloudinary.com/aptidude/image/upload/v1756660696/aptidude_questions/zysqkrmeofvzmymgtllf.png)

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 rings form an equilateral triangle with side length equal to the diameter of the rings, i.e., $21 	ext{ cm}$.

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

4. Calculate the area of one sector of the circle formed by the angle $60^\circ$ (since the triangle is equilateral, each sector angle is $\frac{360^\circ}{6} = 60^\circ$):
   $$	ext{Area}_{sector} = \frac{1}{6} 	imes \pi 	imes r^2 = \frac{1}{6} 	imes \frac{22}{7} 	imes (10.5)^2 = \frac{1}{6} 	imes \frac{22}{7} 	imes 110.25 = 57.75 	ext{ cm}^2$$

5. Total area of three such sectors:
   $$3 	imes 57.75 = 173.25 	ext{ cm}^2$$

6. Area enclosed by the three rings (the space between the rings) is the area of the triangle minus the area of the three sectors:
   $$	ext{Enclosed Area} = 190.953 - 173.25 = 17.703 \approx 17.70 	ext{ cm}^2$$

**Answer:** $17.70 	ext{ cm}^2$ (Option 2)

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