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# A pair of the identical circles overlap each other so that the distance between their centers is equql to their mutual radius, which ha

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A pair of the identical circles overlap each other so that the distance between their centers is equql to their mutual radius, which has a value of 10 cm. Inscribe a square in a middle region, and find the area of the same.

Guest Jun 6, 2015

#1
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Here's a graph of a possible orientation........https://www.desmos.com/calculator/k0yljiphsh

The center of this square will be at the origin.......and the diagonals of the square will lie on the lines y = x  and y = -x.  And finding the intersections of these lines with these circles wil give us the vertices of the square.

And the four vertices of the square will be located at:

(5(√7 - 1)/2,  5(√7 - 1)/2), (5(√7 - 1)/2, - 5(√7 - 1)/2), (-5(√7 - 1)/2, - 5(√7 - 1)/2), (-5(√7 - 1)/2,  5(√7 - 1)/2)

And the area will be ( 5(√7 - 1))^2  =  about 67.712 sq units

CPhill  Jun 6, 2015
#1
+93038
+13

Here's a graph of a possible orientation........https://www.desmos.com/calculator/k0yljiphsh

The center of this square will be at the origin.......and the diagonals of the square will lie on the lines y = x  and y = -x.  And finding the intersections of these lines with these circles wil give us the vertices of the square.

And the four vertices of the square will be located at:

(5(√7 - 1)/2,  5(√7 - 1)/2), (5(√7 - 1)/2, - 5(√7 - 1)/2), (-5(√7 - 1)/2, - 5(√7 - 1)/2), (-5(√7 - 1)/2,  5(√7 - 1)/2)

And the area will be ( 5(√7 - 1))^2  =  about 67.712 sq units

CPhill  Jun 6, 2015
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I had a lot of fun solving this one using trigonometry. It's a chalange -- but, it can be done. Thanks Anon!

Guest Jun 6, 2015
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oops, I misspelled   "challenge"

Guest Jun 6, 2015
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Good thinking Chris !

It would have been nice if you had given us the jist of your logic anon :)

Melody  Jun 6, 2015
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Thanks, Melody.....!!!

CPhill  Jun 6, 2015
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Melody: "It would have been nice if you had given us the jist of your logic anon :)"

I'll try to explain what I did; I'll use a clock (hour hand).

I set the centers of the circles (A and B) at 9:00 and 3:00, then I added 2 more centers (C and D) at 12:00 and 6:00. The circumferences of these circles have created 4 vertices of the square (W, X ,Y and Z) at 10:30, 1:30, 4:30 and 7:30. I drew the sides and the diagonals of the square, and marked the center of it with an (O).

If you draw an isoceles triangle (BDW), and mark a midpoint of (BD) with an (M), we get a perfect right triangle (BMW).

BW = 10cm,   BO = 5cm,   BM = sin45o*5 = 3.54cm,   BM = MO,   MW = 9.35cm,  OW = MW-MO = 5.82cm

The diagonal of the square:   2OW = 11.62cm

The side of the square:    WX = sin45o*11.62 = 8.22cm

(And finally), the area of the square (WXYZ) is:    67.71cm2

It would be nice if someone made a graph of all this; if you decided to do it, do not draw the whole circumferences; do them to the point where they cross each other; it's gonna look nicer. Thanks!

I'm  exhausted !!!

Guest Jun 6, 2015
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+27251
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Here's yet another way of looking at the problem:

.

Alan  Jun 6, 2015