geometry
The area of a square is $a^2$. Two quarter circles intersect in the square and form a symmetry along the square’s diameter. Where have I failed in the process of finding the area of the intersection?
Thank you!
From the lower left-hand corner of the rectangle (let's call that point $A$), consider the diagonal line as a secant line of the left-hand circle, intersecting the circle at points $B$ and $C$, where $B$ is between $A$ and $C$. Also consider one of the edges of the rectangle adjacent to $A$ as a tangent line touching the circle at $D$.
Then by a theorem about tangent and secant lines from a point outside a circle, we have the following relationship of the lengths of the segments from $A$ to each of the three points $B$, $C$, and $D$: $$ (AD)^2 = AB \times AC. \tag1 $$
It is easy to find that $AD = \frac12a$. Now let $E$ be the midpoint of the bottom side of the rectangle; then $AC$ is the hypotenuse of right triangle $\triangle AEC$, which has legs $a$ and $\frac12a$, and therefore $AC = (\frac12\sqrt5)a$.
We can then use Equation $(1)$ to find the length $AB$, so we can find the length of the chord $BC$; from that chord and the radius of the circle we can get the angle of $S_1$ at the center of the circle.
Firstly, to remove any doubts, you are correct so far in that:
However, you forget that they overlap. This means that while the quarter circles sum to $50\pi$ in area, some of it just doesn't matter, since the other circle is already covering up that bit of space. (If you've ever heard of the inclusion-exclusion principle for stuff like probabilities and counting problems, it has a very similar feel to it.) Consider the below picture, which only has one of the circles in it. Imagine rotating a copy of this square $180^\circ$ and superimposing it on itself: you obviously end up with the original picture you have, but with overlap:
In fact, it is the area of the overlap you are tasked to find. That overlap is the blue area in the picture below:
With the picture color-coded like this, it should not be difficult to convince you that
$$\text{(The blue area)} = \text{(The square's area)} - \text{(The two bits of green area)}$$
In a similar vein, the first picture should show you that the area of one portion of the green area can be obtained by taking the square, and subtracting the area of the quarter circle. Double that for the green area, and subtract that from the area of the square. Then you have your result, the area of the overlapping region.
Best Answer
You have successfully found the area of the quarter circle to be $$\frac 14 \pi a^2$$
We can say that the area we are trying to find is called $A$ and we can also see that the two quarter circles cover the whole area of the square and overlap in the area $A$. Therefore \begin{align}2\times \text{quarter circle area} &= \text{sqaure area} + \text{desired area}\\ 2\times \left(\frac14 \pi a^2\right)&=a^2+A\\ A&=2\times \left(\frac14 \pi a^2\right)-a^2\\ &=\frac12 \pi a^2-a^2\\ &=a^2\left(\frac \pi2-1\right)\\ &\approx 0.571a^2\end{align}
That is to say that the desired area is around $57\%$ the size of the square