Question

Three circles with radii 1 cm each are drawn touching each other. Find the area of the shaded portion (in cm$^2$). \includegraphics[width=0.5\linewidth]{image85.png}

• A. $\dfrac{\sqrt{3}-\pi}{2}$
• B. $\dfrac{2\sqrt{3}-\pi}{2}$
• C. $\dfrac{\pi-\sqrt{3}}{2}$
• D. $\dfrac{\pi-2\sqrt{3}}{2}$

Hint:

For “gaps” between equal tangent circles, connect centres to form an equilateral triangle and subtract the matching circular sectors (\(60^\circ\) each).

Answer 1

The Correct Option is B

Step 1: Join the centres.
Since all three circles have radius \(r=1\) cm and touch pairwise, the segment between any two centres equals \(2r=2\) cm. Hence the three centres form an equilateral triangle of side \(2\).
Step 2: Area of the equilateral triangle.
For side \(a=2\):\quad \(A_{\triangle}=\dfrac{\sqrt{3}}{4}a^{2}=\dfrac{\sqrt{3}}{4}\times 4=\sqrt{3}\ \text{cm}^2.\)
Step 3: Subtract the three $60^\circ$ sectors.
At each vertex, the angle is \(60^\circ\), so each circular sector has area \(\displaystyle A_{\text{sector}}=\frac{60^\circ}{360^\circ}\pi r^2=\frac{1}{6}\pi(1)^2=\frac{\pi}{6}\ \text{cm}^2.\)
There are three such sectors - total removed area \(\displaystyle 3\times\frac{\pi}{6}=\frac{\pi}{2}\ \text{cm}^2.\)
Step 4: Shaded area.
\(\displaystyle A_{\text{shaded}}=A_{\triangle}-A_{\text{sectors}} =\sqrt{3}-\frac{\pi}{2} =\frac{2\sqrt{3}-\pi}{2}\ \text{cm}^2.\)
\[ \boxed{\dfrac{2\sqrt{3}-\pi}{2}\ \text{cm}^2} \]