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?

Based on the question, the following figure is obtained. The shaded area denotes the area that we need to find. 

Here, area of the equilateral triangle is $$A=\frac{\sqrt{\ 3}}{4}\left(side\right)^2=\frac{\sqrt{\ 3}}{4}\times\ 24^2=144\sqrt{\ 3}sq\ units$$

The inradius of the circle will be given by $$r=\frac{A}{s}$$ where A is the area of the triangle and s is the semi perimeter which is found to be $$\frac{24+24+24}{2}=36$$

So, $$r=\frac{144\sqrt{\ 3}}{36}=4\sqrt{\ 3}units$$

Hence, area of the circle is $$\pi\ \times\ \left(4\sqrt{\ 3}\right)^2=48\pi\ \ sq\ units$$

Finally, the area of the remaining area will be teh difference, which is $$144\surd{3} - 48 \pi cm^{2}$$ sq units. Hence , option A is the correct choice.