Three equal circles are described with vertices of the triangles as centres. If the radius of each circle is $r$, the sum of areas of the portions of the circles intercepted in a triangle is (Campus Recuritment, 2010)

### Concept & Area of Sectors in a Triangle The sum of the interior angles of any triangle is always $180^\circ$ (or $\pi$ radians). The intercepted portions of the circles at each vertex represent three sectors of a circle with a common radius $r$. $$\text{Area of a sector} = \frac{\theta}{360^\circ} \times \pi r^2$$ ### Step-by-Step Solution * **Given:** Three circles with radius $r$ centered at the vertices of a triangle. * **Deduction:** Let the interior angles of the triangle be $\theta_1$, $\theta_2$, and $\theta_3$. We know that $\theta_1 + \theta_2 + \theta_3 = 180^\circ$. * **Calculation:** The area of the sector at the first vertex is $\frac{\theta_1}{360^\circ} \times \pi r^2$. * **Calculation:** The total area of the three sectors is $\frac{\theta_1 + \theta_2 + \theta_3}{360^\circ} \times \pi r^2$. * **Calculation:** Substitute the sum of angles: $\frac{180^\circ}{360^\circ} \times \pi r^2$. * **Calculation:** Simplify the fraction: $\frac{1}{2}\pi r^2$. ### Exam Strategy & Shortcut Visualize bringing the three corners of any triangle together. Because their angles sum to $180^\circ$, joining the three sectors perfectly forms a semi-circle. The area of a semi-circle with radius $r$ is simply $\frac{1}{2}\pi r^2$. ### Common Pitfall Students often try to assume the triangle must be equilateral and calculate for three $60^\circ$ sectors individually, which wastes time (though it yields the same correct answer). The principle works universally for any triangle. ### Final Answer Therefore, the correct answer is **$\frac{1}{2}\pi r^2$**.