A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of the remaining portion of the triangle?

For an equilateral triangle with inscribed circle:

$\text{Side } = a = 24 \text{ cm}$

$\text{Area of equilateral triangle} = \frac{\sqrt{3}}{4} \times a^2$

$= \frac{\sqrt{3}}{4} \times 24^2$

$= \frac{\sqrt{3}}{4} \times 576$

$= 144\sqrt{3} \text{ cm}^2$

$\text{Radius of inscribed circle} = \frac{a}{\sqrt{12}} = \frac{24}{\sqrt{12}} = 4\sqrt{3} \text{ cm}$

$\text{Area of inscribed circle} = \pi r^2$

$= \pi(4\sqrt{3})^2$

$= 48\pi \text{ cm}^2$

$\text{Remaining area} = \text{Area of triangle} - \text{Area of circle}$

$= 144\sqrt{3} - 48\pi \text{ cm}^2$

Therefore, the answer is $144\sqrt{3} - 48\pi \text{ cm}^2$