Question: Intersecting Circles

Comment on Intersecting Circles

"Entire Geometry module (including practice questions) in 1 slideshow" file was very helpful. Especially for fast review. It will be better if other modules also have such that.
greenlight-admin's picture

Thanks for that feedback. We'll definitely consider that for the future.

we can use formula for above question

(4 ( 1/6 pi r^2) ) - (sqrt(3)/2 * r^2)

will break down the formula,

4 * (area of sector formed by radii with inscribed angle 60) - area of rhombus

area of sector will be 60/360 * pi * r^2 => 1/6 * pi * r^2

area of rhombus : one of the diagonal is r itself and other will be sqrt(3) * r, therefore area = ( sqrt(3) * r * r) ) / 2

coming to the problem r = 6.

24 pi => area of 4 sectors, redundantly we have added rhombus area. We need subtract 18 sqrt(3).

A is the answer.
greenlight-admin's picture

Sorry, I'm not sure what the following formula represents: (4 ( 1/6 pi r^2) ) - (sqrt(3)/2 * r^2)

That formula represents the shaded region. this is true only for the case in which, two circle with same radius having their center on circumference of each other.

In that formula, first part represent the area of sector formed by 60 degree angle and 2 radius. We have 4 sectors as such.

while calculating the area of sector, we have redundantly added rhombus kind of region. we need to eliminate it. so will subtract the rhombus region from 4 times area of sector, this is what second region represents.

(4 * (60/360) pi r^2) - ( ( (sqrt(3)) / 2 ) * r^2 )

I remember in one of the video, some student has asked if any easy solution possible, since it is lengthy. Thought deriving this formula is quite easy. Don't know whether it is convincing.
greenlight-admin's picture

Okay, I see. That formula certainly works, but it could be a pretty cumbersome one to memorize :-)

How to calculate another diagonal of rhombus? How did you get sqrt (3)*r?please explain?
greenlight-admin's picture

How to calculate another diagonal of rhombus?
Are you referring to the diagonal of the rhombus in the video question? The horizontal diagonal is equal to the radius of each circle. So, that diagonal has length 6.

How did you get sqrt(3)*r?
Are you referring to the area of one of the equilateral triangles?
If so, the area of an EQUILATERAL triangle = (√3)(side²)/4
So, I just plugged in the side length, which is 6.

Got it. Thank you.

@ 2.15
As the sectors are added:
(a+b) + (d + e) + (b + c) + (d + f)
so won't the sector b and sector d counted twice?
greenlight-admin's picture

Yes, we count sectors b and d twice.
At 2:20 in the video, we deal with that by subtracting b and d from from both sides of the equation.



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