Three circles each of radius 3.5 cm are drawn so that each touches the other two. Find the area enclosed between these circles.
\(\dfrac{49\sqrt{3}}{4} - \dfrac{49\pi}{8}\,\text{cm}^2 \;\approx\; 1.97\,\text{cm}^2\)
Step 1: Mark the centers of the three circles as \(A, B, C.\) Since each circle touches the other two externally, the distance between any two centers is equal to twice the radius.
So, \(AB = BC = CA = 2r = 2 \times 3.5 = 7\,\text{cm}.\)
Step 2: The three centers form an equilateral triangle of side \(7\,\text{cm}.\)
Step 3: Area of an equilateral triangle of side \(a\) is:
\[ A_{\triangle} = \dfrac{\sqrt{3}}{4} a^2 \]
Substitute \(a = 7\,\text{cm}:\)
\[ A_{\triangle} = \dfrac{\sqrt{3}}{4} (7)^2 = \dfrac{49\sqrt{3}}{4}\,\text{cm}^2. \]
Step 4: Now look at the circular parts inside the triangle. At each corner, there is a sector of angle \(60^\circ\) (because angles of an equilateral triangle are \(60^\circ\)).
Area of one sector of radius \(r = 3.5\,\text{cm}\) and angle \(60^\circ\) is:
\[ A_{\text{sector}} = \dfrac{60}{360}\pi r^2 = \dfrac{1}{6} \pi (3.5)^2 \]
\[ A_{\text{sector}} = \dfrac{1}{6} \pi (12.25) = \dfrac{12.25\pi}{6}\,\text{cm}^2. \]
Since there are 3 such sectors, total area of sectors is:
\[ A_{\text{sectors}} = 3 \times \dfrac{12.25\pi}{6} = \dfrac{36.75\pi}{6} = 6.125\pi = \dfrac{49\pi}{8}\,\text{cm}^2. \]
Step 5: The required shaded area (enclosed between the circles) is the triangle area minus the three sectors:
\[ A = A_{\triangle} - A_{\text{sectors}} \]
\[ A = \dfrac{49\sqrt{3}}{4} - \dfrac{49\pi}{8} \;\,\text{cm}^2. \]
Step 6: Approximate value:
\( \dfrac{49\sqrt{3}}{4} \approx 21.22\,\text{cm}^2, \quad \dfrac{49\pi}{8} \approx 19.25\,\text{cm}^2. \)
So, \( A \approx 21.22 - 19.25 = 1.97\,\text{cm}^2. \)
Final Answer: The required area = \(1.97\,\text{cm}^2\).