The midpoint of half circle HC1 will sit somewhere on the horizontal leg providing the arc is always contained within the Area of the Quadrant.
1. For the purpose of this exercise, the radius of the HC1 just happens to be 3m which puts it at the midpoint of the Quadrant horizontal radius.
2. The radius of HC1 can never be bigger than half the radius of the Quadrant (6m)

The midpoint of half circle HC2 will sit somewhere on the vertical leg providing the arc is always contained within the Area of the Quadrant.
1. For the purpose of this exercise, the radius of the half circle HC2 is unknown.
2. Given that the radius of HC1 is 3m, the radius of HC2 will need to be smaller than the radius of HC1 otherwise the half circles will overlap.
3. The half circle HC2's arc start point has been given as the very top of the vertical leg of the quadrant.
4. For the purpose of this exercise, half circles HC1 & HC2 will be tangent (externally) to each other.

Neither of the half circles shall overlap one another.

Neither of the half circles shall extend past the outline of the Quadrant.

Since we know that the radius of the Quadrant is 6m, its area can be calculated by using the formula =(3.14159265358979*6^2)/4 which results in 28.27433388 sq mtrs.

Since we also know that the radius of HC1 is 3m, its area can be calculated by the formula =(3.14159265358979*3)/2 which results in 14.13716694 sq mtrs.

We need to calculate the radius of the half circle HC2 in order to determine the area of HC2. By drawing a line from the midpoint of HC1 on the horizontal leg of the Quadrant to the midpoint of the half circle HC2 on the vertical leg of the Quadrant, we are effectively defining the length of the hypotenuse of a triangle formed by the points Mid point HC1, midpoint HC2 and the joining point of the horizontal and vertical legs of the Quadrant.

From this we can deduce that the following is true:
1. The length of the horizontal leg is 3m.
2. The length of the vertical leg is the radius of HC2 (r2) - 6m.
3. The length of the hypotenuse is the radius of both HC1(R1) & HC2 (R2).
Therefore: (3+(R2-6)) squared = (3 + R2)squared
Which can be written as (3^2)+ (R2^2 + 6R2 + 6R2 + 36) = (3^2 + R2^2) or 9 + R2^2 +12R^2 - 36 = 9 + R2^2
Which results in R2 = 2.
Therefore the following may be deduced;
1. The length of the base is 3m,
2. The length of the vertical leg is 4m.
3. The length of the hypotenuse is 5m.
4. Area of HC2 is 6.283185306 sq mtrs.

Thus the area remaining in the quadrant is 28.27433388 - 14.13716694 - 6.283185306. = 7.853981634 sq mtrs.

Since we are having difficulty in creating a Function, can somebody confirm that my maths is correct and is there a way to write to correct formulas into excel. Once I'm able to move past this point then I shall concentrate on a Function to do the same thing.

Hopefully this clears up any confusion caused by the previous posts.