Let ABC be an equilateral triangle, with each side of length k. If a circle is drawn with diameter AB, then the area of the portion of the triangle lying inside the circle is

ABC is an equilateral triangle with side length k , and a circle is drawn with AB as the diameter. So the radius of the circle is $$\dfrac{k}{2}$$

Let D and E be the points where the circle and triangle intersect. O is the centre of the circle. The following figure represents the given situation.

$$\angle\ ABC=\angle\ ACB=\angle\ BAC=60^{\circ\ }$$

Also, OA=OB=OD=OE as all of them are radii of the same circle. So, this means, for $$\triangle\ OAD,\ \angle\ OAD=\angle\ ODA=60^{\circ\ }$$, as the sides are equal. By angle sum, we can say that $$\angle\ DOA=180-60-60=60^{\circ\ }$$. So, AOD is also equilateral with side length $$\dfrac{k}{2}$$ 

Similarly we can also conclude that BOE is also equilateral with side length $$\frac{k}{2}$$

Again, AOB is a straight line, so $$\angle\ AOD+\angle\ DOE+\angle\ EOB=180^{\circ\ }$$, which gives us $$\angle\ DOE=180-60-60=60^{\circ\ }$$

Now, we are asked to find the area of the portion of the triangle lying inside the circle, which is the sum of the areas of the 2 triangles AOD and COB and the sector DOE of the circle. 

Now, 

$$ar\triangle\ AOD=\dfrac{\sqrt{\ 3}}{4}\left(side\right)^2=\dfrac{\sqrt{\ 3}}{4}\left(\dfrac{k}{2}\right)^2=\dfrac{\sqrt{\ 3}k^2}{16}sq\ units$$

which is also the area of $$\triangle\ BOE$$

And fo the sector DOE, area = $$\dfrac{60}{360}\times\ \pi\ r^2=\dfrac{1}{6}\times\ \pi\ \left(\dfrac{k}{2}\right)^2=\dfrac{\pi k^2}{24}\ sq\ units$$

So the required total area = $$\dfrac{\sqrt{\ 3}k^2}{16}+\dfrac{\pi k^2}{24}+\dfrac{\sqrt{\ 3}k^2}{16}=2\dfrac{\sqrt{\ 3}k^2}{16}+\dfrac{\pi k^2}{24}=\dfrac{\left(3\sqrt{\ 3}+\pi\ \right)}{24}k^2$$ , which is option A