An equilateral triangle has been drawn inside the circle. Split the triangle to two parts with equal area by a line segment parallel to the triangle side. Draw an inscribed circle inside this smaller triangle. What is the ratio of the area of this circle compared to the area of original circle.
Let the radius of the larger circle be $R$. Side-length of the larger equilateral triangle is $\sqrt{3}R$. If the side-length of the smaller equilateral triangle be $a$, we have $\frac{\sqrt{3}.3R^2}{4}=\frac{1}{2}.\frac{\sqrt{3}a^2}{4}\implies a=\sqrt{1.5}R$. Using $r=\frac{\Delta}{s}$, we have the radius of the incircle to be $r=\frac{R}{2\sqrt{2}}$. So ratio of areas is $1:8$.