From a cirlce of radius $a$, an isosceles right angled triangle with the hypotenuse as the diameter of the circle is removed. The distance of the centre of gravity of the remaining position from the centre of the circle is :

The centroid of the semicircular area of radius a lies at a distance of 4a/(3*pi) from the center along the axis of symmetry. The right-angled isosceles triangle removed has its hypotenuse as the diameter, meaning its altitude is a, and its centroid lies at a distance of a/3 from the center along the same axis. Taking moments about the center for the remaining area, (pi*a^2/2)(4a/(3*pi)) - (a^2)(a/3) = ((pi*a^2/2) - a^2)x. Simplifying the left side gives (2a^3/3) - (a^3/3) = a^3/3. Equating this to the right side and solving for x results in x = a / (3(pi - 1)).

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