Three equal circles are drawn on a triangle ABC, with points A, B and C as the centres. The radius of each of the circle is equal to half of the side of the triangle ABC. I. The area of the shaded region 2 (in cm²) II. 30 cm²

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Step 1: Let the side of the (equilateral) triangle be $s$ and the common circle radius be $r$. Given the radius equals half the side, $r=\dfrac{s}{2}$, so $s=2r$. The triangle is equilateral so each vertex angle is $60^\circ$.

Step 2: Area of triangle ABC = $\dfrac{\sqrt3}{4}s^2=\dfrac{\sqrt3}{4}(2r)^2=\sqrt3\,r^2$.

Step 3: Each circular sector (centered at a vertex, angle $60^\circ$) has area $\dfrac{60}{360}\pi r^2=\dfrac{1}{6}\pi r^2$. Three such sectors have total area $3\cdot\dfrac{1}{6}\pi r^2=\dfrac{1}{2}\pi r^2$.

Step 4: The shaded region 2 (area inside the triangle but outside the three sectors) is
$$I=\text{triangle area} - \text{sectors area}=\sqrt3\,r^2 - \tfrac{1}{2}\pi r^2 = r^2\Big(\sqrt3 - \tfrac{\pi}{2}\Big).$$
Numerically $\sqrt3-\dfrac{\pi}{2}\approx 0.16125$, so $I\approx 0.16125\,r^2$.

Step 5: To have $I\ge 30$ would require $r^2\ge \dfrac{30}{0.16125}\approx 186.17$, i.e. $r\gtrsim13.64$ (so side $s=2r\gtrsim27.28$). For the given triangle configuration (with ordinary diagram-scale dimensions) $r$ is much smaller, therefore the computed shaded area satisfies
$$I<30.$$
Hence Quantity I < Quantity II, so Quantity I < Quantity II.

FINAL_ANSWER: c

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