Compare the areas

First observe that the area of the quarter circle with centre B is made up of

area X + the area of the two semi circles - area Y

The area Y needs to be taken away because it has been counted twice, once in each semi circle.

Since we are looking for a ratio we can choose any scale we like to measure the problem. To ease the calculations suppose the square has a side of 2 units.

Then the big quarter circle has area $\dfrac{1}{4}\pi2^2=\pi$

Each semi circle has area $\dfrac{1}{2}\pi1^2=\dfrac{\pi}{2}$

So by the above observation

$\pi=x+\pi-y$ and so

$x=y$

and so the required ratio is $\boxed{1}$

Let the center of the square be O

Let the area of the region OCB = a

the area of the region OCB = the area of the region OAB = a

Let the length of the side of the square be 2r

a + y =(1/2)(Pi) (r^2).............................. (1)

a + x = (1/4)(Pi)[(2 r)^2] -(1/2)(Pi) (r^2) =(1/2) (Pi) (r^2) .......................... (2)

From (1), (2) we get

x = y

So

x : y = 1 : 1

Let's call X=S1 and Y=S2

Let the center of the square be $O$ , the midpoints of $BC$ and $AB$ be $P$ and $Q$ respectively and the side length of the square be $a$ . We note that:

$y = 2(\text{Area of Quadrant }QOB-\text{Area of }\triangle QOB)$

$\quad = 2 \left[ \dfrac {1}{4} \left( \dfrac {\pi a}{2} \right)^2 - \dfrac {1}{2} \dot{} \dfrac {a}{2} \dot{} \dfrac {a}{2} \right] = \dfrac {\pi a^2}{8} - \dfrac {a^2}{4}$

$x = \text{Area of Quadrant }ACB-\text{Area of Semicircle }AOB - \\ \quad \quad \text{Area of Semicircle }COB + y$

$\quad = \dfrac {1}{4} \dot{} \pi a^2 - 2 \left( \dfrac {1}{2} \dot{} \dfrac {\pi a^2}{4} \right) + \dfrac {\pi a^2}{8} - \dfrac {a^2}{4} = \dfrac {\pi a^2}{8} - \dfrac {a^2}{4} = y$

Therefore, $\dfrac {x}{y} = \boxed{1}$

The area of the quadrant $ABC$ minus the areas of semicircles with diameters $AB$ and $BC$ gives us $x - y$ . But this difference of areas equals $\frac {(\pi) (a)^2 }{4} - 2* \frac {(\pi) (a/2)^2 }{2} = 0$ . Hence $x -y =0$ , and $x:y = 1$ .

The square ABCD has side "a". There are two semi-circles of radius $\frac {a}{2}$ and area $S_{1}$ each and a quarter of a circle of radius "a" and area $S_{2}$ . Hence:

1. $S_{1}$ = $\pi \times \frac {a^{2}}{8}$

2. $S_{2}$ = $\pi \times \frac {a^{2}}{4}$

Thus, observing the areas in the figure we can affirm that:

X = $S_{2}$ - $S_{1}$ - ( $S_{1}$ - Y)

Hence:

X = $\pi\frac {a^{2}}{4}$ - $2\pi\frac {a^{2}}{8}$ + Y

X = $\pi\frac {a^{2}}{4}$ - $\pi\frac {a^{2}}{4}$ + Y

X = 0 + Y

X = Y Then $\frac {X}{Y} = 1$

If we analyse the two areas, we can see that

Area X = Area of Quadrant - [Sum of Area of Two Semi-circles] + Area Y )

Let

$r = AB= BC = CD = DA$ be the length of the side of square ABCD.

We can also see that r is the length of the radius of the circle where the quadrant ABC is derived and r is also the diameter of a circle where the two semicircles are derived.

$Area X = \frac {1}{4} \pi AB^{2} - [ \frac{1}{2} \pi (\frac{AB}{2})^{2} + \frac{1}{2} \pi (\frac{BC}{2})^{2}] + Area Y$

$Area X = \frac {1}{4} \pi r^2 - [\frac{1}{2} \pi (\frac{r}{2})^{2} + \frac{1}{2} \pi (\frac{r}{2})^{2}] + Area Y$

$Area X = \frac{1}{4} \pi r^2 - [\frac{1}{2} \pi \frac {r^{2}}{4} + \frac{1}{2} \pi \frac {r^{2}}{4}] + Area Y$

$Area X = \frac {1}{4} \pi r^2 - \frac{1}{4} \pi r^{2} + Area Y$

$Area X = Area Y$

So the ratio of the two areas is 1.0.

To calculate ${x}$ you find the area of the large quarter - circle and subtract the area of the overlapping smaller semi-circles. Of course if we just call the area of the two semi-circle $\pi$ $r^{2}$ then we have counted ${y}$ twice. So the area must be $\pi$ $r^{2}$ - ${y}$ . $\therefore\ x = \pi\ r^2 - (\pi\ r^2 - y) = y \Rightarrow\frac{x}{y} = 1$