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## Comment on

Intersecting Circles## "Entire Geometry module

## Thanks for that feedback. We

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

## we can use formula for above

(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.

## Sorry, I'm not sure what the

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

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.

## Okay, I see. That formula

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

## How to calculate another

## How to calculate another

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.

## Hi,

@ 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?

## Yes, we count sectors b and d

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.

Cheers,

Brent

## Thanks