We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
62
4
avatar

Which is larger, the blue area or the orange area?

 

 Nov 14, 2019
 #1
avatar+105411 
+2

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   !!!!

 

 

cool cool cool

 Nov 14, 2019
 #2
avatar+70 
+2

 

 

 

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.

 Nov 14, 2019
 #3
avatar+105411 
+1

Thanks, Gadfly....I like that approach  !!!

 

 

cool cool cool

 Nov 14, 2019
 #4
avatar+23575 
+2

Which is larger, the blue area or the orange area?

\(\begin{array}{|rcll|} \hline {\color{orange}\text{orange}} +3\times {\color{blue}\text{blue}} &=& \pi r_{\text{circumcircle}}^2 \quad | \quad : {\color{orange}\text{orange}} \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{ \pi r_{\text{circumcircle}}^2 } { {\color{orange}\text{orange}} } \quad | \quad {\color{orange}\text{orange}} = \pi r_{\text{incircle}}^2 \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{ \pi r_{\text{circumcircle}}^2 } { \pi r_{\text{incircle}}^2 } \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 \\ 3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 -1 \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\left( \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 -1 \right) \\ \\ && \boxed{\text{here, if triangle is equilateral: }\\ \mathbf{2\times r_{\text{incircle}} = r_{\text{circumcircle}}!!!} } \\\\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\left( \left( \dfrac{ 2\times r_{\text{incircle}} } { r_{\text{incircle}} } \right)^2 -1 \right)\\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\times(2^2 -1) \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\times(3) \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& 1 \\ \mathbf{ {\color{blue}\text{blue}} } &=& \mathbf{ {\color{orange}\text{orange}} } \\ \hline \end{array}\)

 

laugh

 Nov 15, 2019
edited by heureka  Nov 15, 2019

5 Online Users

avatar