The Quadrant

Let $r$ be radius of yellow circle, $A_T$ be total area, $A_B$ and $A_R$ be area of blue and red area respectively, and $C$ be area of a complete yellow circle, then: $\begin{aligned} A_T &= \pi(2r)^{2}/4 \\ &= \pi r^{2} \\ A_T &= C \\ \end{aligned}$ Using the other way to compute $A_T$ : $\begin{aligned} A_T &= (C - A_R) + A_B \\ 0 &= -A_R + A_B \\ A_R &= A_B \\ \frac{A_R}{A_B} &= \boxed{1} \end{aligned}$

image in the figure above let AD=k and hence AB=2k

First , area of ADFE = $k^2$ , area of sectors AFE , DFE , DBF , CEF = $\frac{\pi k^2}{4}$ , area of ABC = $\frac {\pi(2k)^2}{4}=\pi k^2$

the area of the darkest region (red in question ) will be (area of sector AFE -( area of ADFE - area of sector DFE)) & area of fainter region{blue in question} will be (area of quadrant ACB -(area of ADFE + area of sector DBF + area of sector CEF)) $\frac{\text{area of darkest region}}{\text{area of fainter region}}=\frac{\frac{\pi k^2}{4}-(1-\frac{\pi k^2}{4})}{\pi k^2-(1+\frac{\pi k^2}{4}+\frac{\pi k^2}{4})}$ $\Rightarrow \frac{\frac{\pi k^2}{2}-1}{\pi k^2-\frac{\pi k^2}{2}-1}$ $\Rightarrow \frac{\frac{\pi k^2}{2}-1}{\frac{\pi k^2}{2}-1}$ $\Rightarrow \boxed{1}$

The large circle whose quadrant we see has diameter twice that of the two semicircles. In general, a circle with radius R has 1/4 the area of a circle with radius 2R, therefore the sum of the areas of the semi-circles (which together constitute a circle with radius R) should be equal to the area of the quadrant (1/4 of a circle with radius 2R). The semi-circles overlap in the red region, meaning that the total area of the two semi-circles is yellow + yellow + red + red = whole quadrant = yellow + yellow + red + blue, therefore red = blue.

Intuitive proof:

The complete quadrant has twice the area of each semicircle. (Thus, if the semicircles' area were transformed so that they did not intersect, they would fill the quadrant completely.) It follows that the area where they intersect must be equal to the area which neither of them contain.

sum of the areas of two semicircles is equal to quarter circle.therefore,overlapping area(red) is the left area(blue).therefore ratio is 1:1.As simple as you think.

Assign the radius of the quarter circle of r, and since the centers of the two semi circles lie on the radius of the semi circle, their radius is r/2. To get the area of the red section, divide the section into two parts. It becomes an arc, and to get the arc; A of quarter circle - area triangle, then multiply to 2. You'll get r^2(pi/8 - 1/4). The blue section, construct a square by meeting the two radii of the semi circles at the other end of the red section. Area blue equals area quarter circle - area semicircle - area square; you'll get the same area as to the red section. Thus, their ratio is 1.

Lets call the radius of the semicircles R

The area of the Blue section is equal to the quarter of the large circle minus the two semicircles plus the red section

Blue=pi4R^2/4 -pi2R^2/2+Red

Blue=piR^2-piR^2+Red

Blue=Red

Red/Blue=1

Let the radius of the quadrant be $r$ , Area of the Blue Region be $B$ and Area of the Red Region be $R$ .
So, Area of the quadrant $=\frac{\pi r^2}{4}$
Also, the radius of the two Yellow semi-circles $=\frac{r}{2}$
We can see from the image that,

(Area of the quadrant)-(Area of the Blue Region)=(Area of the two yellow semi-circles taken together)
Also, we can see that the area of the Red Region is considered twice while adding the area of the two yellow semi-circles.
So,
Area of the two Yellow semi-circles taken together = [ $2\times$ Area of 1 Yellow semi-circle]-[Area of the Red region]
So, we get,
(Area of the quadrant)-(Area of the Blue Region)=[ $2\times$ Area of 1 Yellow semi-circle]-[Area of the Red region]
$\Rightarrow (\frac{\pi r^2}{4})- B = (2\times (\frac{\pi r^2}{8})-R$
$\Rightarrow (\frac{\pi r^2}{4})- B = (\frac{\pi r^2}{4})- R$
$\Rightarrow R-B=0$
$\Rightarrow R=B$
So, $R:B=1:1$
or, we can say,
$\boxed {\frac{R}{B} = 1}$

say radius of quadrant 'r' Area of full Quadrant = pi*r^2/4 -------- (A)

area of semi circle which has dia as radius of Quadrant = pi/2 r^2/4 = pi r^2/8 ------- (B)

observe that (A) = 2*(B)

                     (A) - 2*(B)  =  0

Now,

Area of Blue section = Area of Quadrant - 2*area of semi circles + area of red section

                                   = (A)  - 2*(B) + area of red section

------> area of Blue section = area of red section

------> area of red section/area of Blue section = 1

Revised specific:

(1/4) Pi 2^2 = 2 (1/2) Pi 1^2

=> y + r + y + b = y + r + y + r

=> b = r

=> r/ b = 1

Let y, r, b denote the areas of the yellow, red, blue sections, respectively. Let the radius of the quarter circle be 2, so the radius of the 2 semicircles inscribed in the quarter circle is 1. Now

2y+r+b = pi (2^2)/4 = pi and y+r = pi (1^2)/2 = pi/2,

y+b = pi-y-r = pi-pi/2 = pi/2 and y+r = pi/2,

b-r = 0,

r/b = 1.

2y+r+b=pi(r)^2/4; y+r+y+r=pi(r)^2/8+pi(r)^2/8=pi(r)^2/4; subtracting 1st from 2nd we get r-b=0; r=b therefore r:b=1:1= 1

Let Total Area be A. and area by a single yellow half circle be Ay. and by red part be Ar and blue be Ab. Let radius of whole quadrant be 2r.

So, A=PI r r

and Ay=Pi r r/2

Now, A=Ay+Ay-Ar+Ab

So, A=2*Ay-Ar+Ab

But 2*Ay=A, So, Ar=Ab

So, ratio=1

/(pi r^2/4=2 pi*r^2/8-x+y/) where x=area of red & y=area of blue.The radius of inner semi-circle will become "r/2",where "r" is the radius of the quadrant.Thanks.