# The center of the circle coincides with the apex of an equilateral triangle and its radius is 60% of the side

**The center of the circle coincides with the apex of an equilateral triangle and its radius is 60% of the side of the triangle. In what ratio does a circular arc located inside a triangle divide its area?**

Let the length of the side of the triangle be X cm, then the radius of the circle is R = 0.6 * X cm.

Since triangle ABC is equilateral, its internal angles are 60.

Savs = X ^ 2 * Sin60 / 2 = X ^ 2 * √3 / 4 cm2.

Let us determine the area of the sector bounded by the CM arc.

Ssec = π * R ^ 2 * α / 360 = π * 0.36 * X ^ 2 * 60/360 = π * X ^ 2 * 0.06 cm2.

The area of the AKMS figure is equal to: Sacms = Savs – Ssec = X ^ 2 * √3 / 4 – π * X ^ 2 * 0.06 =

X ^ 2 * (√3 / 4 – 0.06 * π) cm2.

Then Ssec / Sacms = π * X ^ 2 * 0.06 / X ^ 2 * (√3 / 4 – 0.06 * π) = (0.1884) / (0.433 – 0.1884) = 0.77.

Answer: The area ratio is 0.77.