A coffee break geometry problem #4

First, we have that

$\lim_{x \rightarrow 1} \frac{3x^2 - 5x + 2}{x^{2} - 1} = \lim_{x \rightarrow 1} \frac{(3x - 2)(x - 1)}{(x + 1)(x - 1)} = \lim_{x \rightarrow 1} \frac{3x - 2}{x + 1} = \frac{1}{2}$ ,

and $\lim_{x \rightarrow 5} \frac{20x - 100}{x^{2} - 25} = \lim_{x \rightarrow 5} \frac{20(x - 5)}{(x + 5)(x - 5)} = \lim_{x \rightarrow 5} \frac{20}{x + 5} = 2$ .

Thus the radius of the smallest circle is $(\frac{1}{2})*2 = 1$ , and so the radii of the medium size and largest circles are $2$ and $4$ , respectively.

Looking at just the lower triangle, its height $h$ will be the sum of the radius of the smallest circle plus the diameters of the two other circles, i.e., $h = 1 + 2*2 + 2*4 = 13$ . Now given the inherent symmetries of the diagram, and since $BG \perp ED$ , we know that both triangles are isosceles right triangles with height $13$ and base length $2*13 = 26$ , and thus each has an area of $(\frac{1}{2})*26*13 = 169$ .

Now to find the area of the shaded region, we need to subtract the areas of two each of the medium and largest circles from the combined area of the two triangles. This results in an area value of

$2*169 - 2*\pi(2^{2} + 4^{2}) = 338 - 40\pi = \boxed{212.336}$

to $3$ decimal places.