Let the radius of the orange circle = r
So......its area = pi * r^2
We can find the side length,s, of the equilateral triangle thusly
tan (30°) = r / [(1/2)s]
1/ √3 = r / [ (1/2)s ]
(1/ √3) ( 1/2)s = r
s = (2√3) r = [ √12 ] r
And the area of the equilateral triangle is ( √3/ 4 ) ([√12]r)^2 = 3√3 r^2
So....the blue area inside the equilateral triangle =
[area of equailateral triangle - area of small circle ] / 3
[√3 r^2 - (1/3) pi r^2 ] = r^2 [ √3 - pi/3] (1)
The radius of the larger circle can be found as
√[ [√3r ]^2 + r^2 ] = √ [ 3r^2 + r^2 ] = 2r
So....the area of the larger circle = pi (2r)^2 = 4pi r^2
So the area between the side of the equilateral triangle and the larger circle is
[Area of larger circle - area of equilateral triangle ] / 3 =
[ 4pi r^2 - 3√3r^2 ] / 3 = r^2 [ (4/3)pi - √3] (2)
So the sum of (1) and (2) is the sum of the blue areas
r^2 [ √3 - pi/3 + (4/3) pi - √3 ] = pi r^2
So.....the orange and blue areas are equal !!!!
Here is a different approach.The arc AC has measure \(\frac{1}{3}\cdot2\pi\) and the angle AOB has measure half of the arc or \(\frac {\pi}{3}\). If the radius of the larger circle is \(R \), then the measure of OB, the radius of the smaller circle, is \(Rcos(\frac{\pi}{3})\)and so the orange region has area \(\pi(Rcos(\frac{\pi}{3}))^2=\frac{1}{4}\pi R^2\) . The blue region, however, has area equal to \(\frac{1}{3}\)of the difference between area of the larger circle and the area of the smaller circle, i.e. \(\frac{1}{3}(\pi R^2-\frac{1}{4}\pi R^2) =\frac{1}{4}\pi R^2\). So the two regions have the same area, each being \(\frac{1}{4}\) of the area of the larger circle.
Which is larger, the blue area or the orange area?
\(\begin{array}{|rcll|} \hline {\color{orange}\text{orange}} +3\times {\color{blue}\text{blue}} &=& \pi r_{\text{circumcircle}}^2 \quad | \quad : {\color{orange}\text{orange}} \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{ \pi r_{\text{circumcircle}}^2 } { {\color{orange}\text{orange}} } \quad | \quad {\color{orange}\text{orange}} = \pi r_{\text{incircle}}^2 \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{ \pi r_{\text{circumcircle}}^2 } { \pi r_{\text{incircle}}^2 } \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 \\ 3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 -1 \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\left( \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 -1 \right) \\ \\ && \boxed{\text{here, if triangle is equilateral: }\\ \mathbf{2\times r_{\text{incircle}} = r_{\text{circumcircle}}!!!} } \\\\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\left( \left( \dfrac{ 2\times r_{\text{incircle}} } { r_{\text{incircle}} } \right)^2 -1 \right)\\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\times(2^2 -1) \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\times(3) \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& 1 \\ \mathbf{ {\color{blue}\text{blue}} } &=& \mathbf{ {\color{orange}\text{orange}} } \\ \hline \end{array}\)