## Semi-circle Lunes

#### Problem

A triangle is formed by connecting the two ends of the diameter of a semi-circle, length `c`, to a point on the circumference. Circles are constructed on the two shorter sides with diameters `a` and `b` respectively, so as to form two lunes (the shaded part).

Find the total area of the two lunes in terms of `a`, `b`, and `c`.

#### Solution

Consider the diagram, with regions identified by the numbers 1 to 6.

In a semi-circle the triangle will be right-angled, so using the Pythagorean Theorem, `a`^{2} + `b`^{2} = `c`^{2}.

Multiplying through by π/8 we get, ½π(`a`/2)^{2} + ½π(`b`/2)^{2} = ½π(`c`/2)^{2}. In other words, if semi-circles are drawn on the sides of a right-angle triangle, the area of the semi-circles on the shorter sides will be equal to area of the semi-circle on the hypotenuse.

That is, (A_{1} + A_{2}) + (A_{3} + A_{4}) = A_{6}; in addition, A_{6} = A_{2} + A_{4} + A_{5}.

A_{1} + A_{2} + A_{3} + A_{4} = A_{2} + A_{4} + A_{5} A_{1} + A_{2} = A_{5}

Hence the area of the two lunes are equal to area of triangle, ½`ab`.