Sorry in advance for not attaching a picture to help explain what I'm talking about, I don't have the reputation necessary to do so yet.
Consider a square with side length $a$ with a circle inside it, with the circumference just touching the sides of the square (i.e. $2r=a$). 4 smaller circles are located in each corner of the square such that their circumferences touch each side of the square and the larger circle.
All the circles are shaded black and the remaining area of the square is shaded red. Determine the percentage area of the square that's shaded red.
I know that to find the remaining percentage area of a single circle inscribed in a square you'd simply use $A_{\text{%}}=100(1-\frac {A_{circle}} {A_{square}})=100(1-\frac{\pi r^2}{a^2})=100(1-\frac{\pi r^2}{4r^2})=100(1-\frac{\pi}{4}) \approx 21.4602 \text{%}$
However, the 4 smaller circles in the corners have really stumped me with this problem. Any help is greatly appreciated.
Best Answer
Let $b$ the radius of the four small circles. Apply the Pythagorean formula to the isosceles right triangle whose hypotenuse connect the centers of a small and the large circle, i.e.
$$(r+b)^2 = (r-b)^2 + (r-b)^2$$
which yields,
$$b = (3-2\sqrt2)r$$
Then, the total area of the four small circles is $\pi(17-12\sqrt2)a^2$.