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August 31, 2014

August 31, 2014

Posted by **DrBob222** on Tuesday, June 25, 2013 at 1:03am.

You have two interlocking circles and the radius of circle B goes through the center of circle A and of course the radius of circle A goes through the center of circle B. The radius of each circle is 30 feet. Now, if you draw a line C from the top center of circle A across to the top center of circle B, the line would be 60 feet long and would leave an area below it created by the line C and part of an arc of circle A and part of an arc of circle B. Please give me the area of that space in square feet or the formula for how it's worked. Thanks.

- Math -
**MathMate**, Tuesday, June 25, 2013 at 7:03amWe will describe the problem as follows.

Two circles A,B of equal radii (r=30') are centred at P, Q, distance r apart.

Points R and S on circles A, B are such that RS form a common tangent to both circles. Hence PRSQ form a square of side r.

Arcs are drawn with centre P and Q, radius r, which intersect at point D inside the square.

Hence Δ PDQ is an equilateral triangle of side r.

The required area bounded by the side RS , arcs RD and DS will be equal to

Area of square PRSQ - Area of ΔADQ - area of sector RPD - Area of sector SQD.

Note that sectors RPD and SQD have centrai angles of (90-60)=30°.

- Math -
**MathMate**, Tuesday, June 25, 2013 at 7:06amTypo corrections:

Area of square PRSQ - Area of ΔPDQ - area of sector RPD - Area of sector SQD.

Note that sectors RPD and SQD have central angles of (90-60)=30°.

- Math -
**DrBob222**, Tuesday, June 25, 2013 at 11:45amMany thanks.

- Math :) -
**MathMate**, Tuesday, June 25, 2013 at 12:27pmYou're welcome!

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