Thread: Sectors inside a square. Need to find the common area.

Hi,
this Q stumped me.
It looks VERY simple but i couldnt solve it .
Please find the diagram below.
Please let me know if you have any ideas about this.
Thanks in advance

By pure geometry, I'm not sure I'd know where to start, but if we are allowed to express it in co-ordinate geometry, it can be evaluated by integral calculus. That's how I'd do it (but I wouldn't enjoy it!) - are you allowed to use calculus here?

We are not.
but i got this Q in MCQ format & so the method used in not of primary importance.
Please go ahead and psot your solution.
And Thanks for the help

Is ABCD meant to be a square? If so, then:

DE is a radius of the circle centered at D. So also is DC.

CE is a radius of the circle centered at C.

Since the quadrilateral is a square, then the circles' radii are equal. Then the triangle DEC is equilateral. In particular, the angle EDC measures 60 degrees. This means that the angle ADE measures 30 degrees, since ADC is right.

By symmetry, the angle CDF measures 30 degrees. Then the angle EDF is 30 degrees.

Find the area of the sector EDF.

Then note the 45-45-90 triangle formed within this sector, when you draw the vertical and horizontal lines through the center, splitting the square into four equal parts. The full area will be the sector, less the area to the left of this 45-45-90 triangle, and then multiplied by 4.

Whatever is the side length of the square, it is the value of the radii of the four circles, R. Draw a line from the corner D to where the vertical and horizontal lines intersect. Let's call that X. The triangle formed by DXF has height R/2, and a base with the same length as one of the legs of the 45-45-90 triangle. You should be able to use some elementary trigonometry to prove that this base has a length of (R/2)(sqrt[3] - 1).

From this, find the area.

well my answer is

[(a^2)(pi-3(sqrt[3]-1))]/3
where a is the length of the side of the square.
since you got it in an MCQ format you may check it first.

Hi Staple,
Thanks !
Your method is very elegant, but i am lost after the 45-90-45 triangle.
I ahve edited the picture, Can you please have a look and tell me why XDF will have a " height R/2, and a base with the same length as one of the legs of the 45-45-90 triangle". It really doesnt get through to me yet. Either I am drawing the figure incorrectly or I am missing some vital point. Can you please look into it.
Thanks.

Originally Posted by nikhil

well my answer is

[(a^2)(pi-3(sqrt[3]-1))]/3
where a is the length of the side of the square.
since you got it in an MCQ format you may check it first.

thanks, but i am actually looking for a method to solve it. Can you please elaborate how you got this result?

Originally Posted by cybermaths

Hi Staple,
Thanks !
Your method is very elegant, but i am lost after the 45-90-45 triangle.
I ahve edited the picture, Can you please have a look and tell me why XDF will have a " height R/2, and a base with the same length as one of the legs of the 45-45-90 triangle". It really doesnt get through to me yet. Either I am drawing the figure incorrectly or I am missing some vital point. Can you please look into it.
Thanks.

Staple, just to add on -
Of course, I am assuming that XEF is the 45-90-45 angle you are talking about ?
Do let me know where I am mistaken.

I used the concept of geometry and integration. here are the steps.

1) find area of region BFGDHEB using geometry (easy)
2) find the area of AGB using integration.
note: for area of AGB consider two congruent circles whose inbetween distance is equal to their radius. then use integration to find the area of the common region. area of AGB will be half of area of this common region.
3)find area AGD (area of quarter circle-area of AGB)
4)find area of DGC(easy)

now use
Z=2(area of BFGDHEB)+4(area of DGC)-a^2
where Z is area of common region.
calculate it and and answer will be obtained.