Two unit circles intersect each other at two points, and they pass through each other's centers as shown above. What is the area of the shaded region?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Relevant wiki: Length and Area - Composite Figures
B a n d C are the centers of our circles. Triangles A B C a n d B C D are equilateral triangles with each angle=60.
We know the mentioned sides that are equal to the radius. With the application of Pythagoras, we get A D = 3 r and angle A B D = 1 2 0 ⟹ the area of the intersection of the two circles(the one highlighted in yellow) is r 2 × ( 3 2 π − 2 3 ) = 2 [ 2 r 2 ( 3 2 π − s i n ( 3 2 π ) ) ] .The area of the two circles is 2 π r 2 .
Thus, the area of one white part or one green part in our problem is π r 2 − r 2 × ( 3 2 π − 2 3 ) = 6 2 π + 3 3 . Keeping in mind that r = 1 . Multiply this area by 2 , we obtain 3 2 π + 3 3 = 6 4 π + 6 3
Notice that when you calculate the area of one unit circle, you added the intersection area once.
Therefore, when you add the area of two unit circles, you added the intersection area twice
I made that same mistake too
Relevant wiki: Circles - Area
A , the distance between the centers of the circle is 1 .
Consider figureConsider figure B , the area of one shaded part is equal to the area of a circle minus the area of two segments of a circle. We need the value of ∠ θ ,
cos θ = 1 2 1 = 2 1
θ = cos − 1 2 1 = 6 0 .
It follows that 2 θ = 1 2 0 .
We need the value of a , by pythagorean theorem
a = 1 − ( 2 1 ) 2 = 4 3 = 2 3
It follows that 2 a = 3
Consider figure C . The area of one segment is 3 6 0 1 2 0 π − 2 1 ( 2 1 ) ( 3 ) = 3 π − 4 3 . So the area of two segments is 2 ( 3 π − 4 3 ) = 3 2 π − 2 3
So the area of one shaded part is π − ( 3 2 π − 2 3 ) = π − 3 2 π + 2 3 = 6 2 π + 3 3 .
Finally, the area of two shaded part is 2 ( 6 2 π + 3 3 ) = 6 4 π + 6 3
In second last step, why didn’t you subtract area of two segments directly from combined area of two circles? It should have given you the same answer right. I actually did but my answer came out different. Please help
The white area in the center, A w , is two circular segments, each with angle 1 2 0 ∘ and radius 1 .
A w = 2 × 2 R 2 ( 1 8 0 α π − sin α ) = 2 × 2 1 ( 1 8 0 1 2 0 π − sin 1 2 0 ∘ ) = 3 2 π − 2 3
Green area on the left is area of circle minus the white area: π − ( 3 2 π − 2 3 ) = 3 π + 2 3
Total green area is double that. 2 × ( 3 π + 2 3 ) = 3 2 π + 3 = 6 4 π + 6 3
OP can you post a solution? I apologize but I'm not sure about the answer. In my opinion the answer should be 6 4 π + 6 3 , and I would like to see where I am wrong. Thanks.
No, you're right, see my solution
And if you have any issues with the solution, do post a report instead in the future
Thanks. I've updated the answer from 6 8 π + 3 3 to 6 4 π + 6 3 . Those who previously attempted this problem will be marked correct.
In future, if you spot any errors with a problem, you can “report” it by selecting "report problem" in the menu. This will notify the problem creator (and eventually staff) who can fix the issues.
Sorry guyyyyyys!
Problem Loading...
Note Loading...
Set Loading...
Relevant wiki: Length and Area - Composite Figures
Notice that lines O A , O P , O B , A P and B P are all radii of the unit circle, meaning they have a length of 1 .
Since all the lines have the same length, we can say that △ O A P and △ O B P are equilateral triangles.
Now, notice that the area of the shaded region is given as below:
Therefore, we need to calculate each component first.
Area of unit circle = π ( 1 ) 2 = π
Area of sector O A P B = 3 6 0 1 2 0 × π ( 1 ) 2 = 3 π
Area of △ O A P = 2 1 ( 1 ) 2 ( sin 6 0 ∘ ) = 2 1 ( 2 3 ) = 4 3
Area of shaded region
= 2 ( Area of unit circle − 2 × Area of sector O A P B + 2 × Area of △ O A P )
= 2 ( π − 2 × 3 π + 2 × 4 3 ) = 2 ( 3 π + 2 3 ) = 3 2 π + 3 = 3 2 π + 3 3 = 6 4 π + 6 3