The shaded region in the diagram below is called an arbelos - Leaving Cert Mathematics - Question b - 2013

For the area of the arbelos:

A_{arbelos} = rac{1}{2} imes ext{(Circumference of semicircle of radius } r_1) - rac{1}{2} ( ext{area of semicircles of radius } r_2 + r_3)

Here, substituting $r_2 = 2$ and $r_3 = 4$:

A_{arbelos} = rac{1}{2} imes ext{(Area of semicircle with radius 6)} - rac{1}{2}( ext{Area of semicircle with radius 2} + ext{Area of semicircle with radius 4})

After simplification, it can be shown that $A_{arbelos} = ext{Area of circle of diameter } k$.

Letting $r_2 = x$, the area of the arbelos can be represented as:

A = rac{1}{2} imes ext{(Area of semicircle with radius 6)} - rac{1}{2} imes ( ext{Area of semicircle with radius } x + ext{Area of semicircle with radius } (6-x))

After simplification, the area can be expressed as:

A = rac{1}{2} imes ext{Area of semicircle with radius 6} - rac{1}{2}(rac{π}{2} x^2) = π(6-x)x

To find the maximum area, we differentiate the area function:

$A = π(6x - x^2)$

Taking the derivative:

rac{dA}{dx} = π(6 - 2x)

Setting rac{dA}{dx} = 0 gives $x = 3$. Thus, substituting back:

$A(3) = π(3)(3) = 9π ext{ cm}^2.$

To prove $RSTC$ is a rectangle:

• Since $AS$ and $FS$ are angles in semicircles, we have:
  • $∠TSR = 90^ ext{o}$
  • $∠TCA = 90^ ext{o}$
  • $∠SRC = 90^ ext{o}$

Thus, $RSTC$ has four right angles, confirming that it is a rectangle.