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# KCC's Quizzes AQQ264 about 3 aligned circles

1. Quote of the week: "Expecting the world to treat you fairly because you are a good person is a little like expecting the bull not to attack you because you are a vegetarian" - Dennis Wholey

2. New quiz AQQ264 about area inside 3 aligned circles

Good luck!

P.S. Please spread those quizzes among colleagues, friends or family members; we want more participants!

• The answer is R2 * (sqrt(3) - PI/3), with R=radius. For R=4cm you get about 11cm2

Let's consider the "left side" of the problem.

Be A and B the centers of the left and central circles and C and D their up and down intersection points. Since AC=CB=AB=R, we have the equilateral triangle T=ABC with side R. The area Y of the circular sector CAD is Y=(1/3)*PI* Rbecause the angle in A is, given the symmetry of the problem, (2/3)*PI.

To calculate the area of the yellow zone, let's call it "X", you have to substract from the central circle area the area 2*Y and 4 times the area of the circular segment defined by the chord AC, let's call it "Z".

So you have X= PI* R2 - 2* Y - 4*Z.

Let's calculate Z. This is equal to the area of the circular sector CBA minus the area of the equilateral triangle T mentioned above, that's Z= PI* R2 /6 - sqrt(3) *  R2/4. (The area of circular sector CBA is PI* R2 /6 because the angle in B is PI/3).

With some algebraic calculation you get the result X= R2 * (sqrt(3) - PI/3)

• The answer is 10.9577 cm2 per the image attached...

• The area of a sector of and angle of  2θ   is   θ r2.  (θ in radians;  area = (2θ / 2π )  π r2 )

The triangle defined by the two points on the circumference delimiting the said sector and the center of the circle is  r2 cosθ sinθ. ( base of 2 r sinθ, height of  r cosθ).

Define an “arc” as the sector less the supported triangle. This arc has an array of r2( θ - cosθ sinθ).

By symmetry, we are looking at the area of a full circle less four times the area of the said arc.

r2 ( π – 4(θ-cosθ sinθ))

We have θ = 60o   (Indeed, consider the three sides of the triangle supporting the sector, all three sides are equal to r, so we have a equilateral triangle). Θ =1.04719… radian

42 (0.68485…)  = 10.95766… cm2

• Answer: Area = 10.96 sq. cm

• The intersection of the three circles can be broken down into four equal segment areas.  One segment each from the left and right circles and two segments from the center circle.  Half of each segment angle is part of an equilateral triangle shown in red below so the segment angles are 120 degrees or 2*PI/3 radians.  The four segment areas equal 39.3078 sq. cm.  The center circle’s area equals 50.2655 sq. cm.  Therefore, the yellow area equals 10.9577 sq. cm.

• if I move the orange shapes according to the blue arrow, then the contents of the yellow part are in the shape of 4 equilateral red triangles - 2 green circular slices.

v_triangle=sqrt(4^2-(4/2)^2) = sqrt(12) = 2*sqrt(3)
S_triangle=1/2*v_triangle*side=1/2*2*sqrt(3)*4 = 4*sqrt(3) cm^2
YellowZone = 4*S_triangle - 2*S_radial_sector = 4*4*sqrt(3) - 2*pi/6*4^2 = 16*(sqrt(3)-pi/3)=10.96 cm^2

• Bingo Gaetano! You get it, big applause!

• Fantastic  , you have found the solution, congratulations!

• YES! You solve the quiz, congratulation  !

• Beautiful  ! You have solved that challenge, big applause!