find area of shaded region; two circles

galactus

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Has anyone ever seen this before?. I seen it in a puzzle book somewhere.

I drew this with paint so it's rather hackneyed, but it should suffice.

Find the area of the shaded region. The circles have radii, 3 and 1 respectively.

circlescg1.gif
 
Hello, galactus!

Find the area of the shaded region. The circles have radii, 3 and 1 respectively.
Code:
              * * *
          *           *
        *               *
       *                 *

      *         P         *
      *         *         *  * *
      *        2|         **     *
                |         *   Q   *
       *       C+--------**---*   *
        *       |       *:*   |1  *
          *    1|     *::::*  |  *
      - - - - * * * - - - - -* * - -
                A             B

Draw PQ\displaystyle PQ. \(\displaystyle \:pQ\,=\,4\)

Right triangle PCQ has side PC=2\displaystyle PC\,=\,2 and hypotenuse PQ=4.\displaystyle PQ = 4.
. . Then CQ=23\displaystyle CQ\,=\,2\sqrt{3}
The triangle is a 30-60 right triangle: QPC=60o,PQB=120o\displaystyle \,\angle QPC\,=\,60^o,\:\angle PQB\,=\,120^o


Trapezoid PQBA has bases 3\displaystyle 3 and 1\displaystyle 1, and height 23\displaystyle 2\sqrt{3}
. . Its area is: 12(23)(3+1)=43\displaystyle \:\frac{1}{2}(2\sqrt{3})(3\,+\,1)\:=\:4\sqrt{3}

Sector QPA\displaystyle QPA in circle P\displaystyle P has area: 12(32)(π3)=3π2\displaystyle \,\frac{1}{2}(3^2)\left(\frac{\pi}{3}\right) \:=\:\frac{3\pi}{2}

Sector PQB\displaystyle PQB in circle Q\displaystyle Q has area: 12(12)(2π3)=π3\displaystyle \,\frac{1}{2}(1^2)\left(\frac{2\pi}{3}\right)\:=\:\frac{\pi}{3}


Therefore, the shaded region has area: 433π2π3  =  4311π6\displaystyle \:4\sqrt{3}\,-\,\frac{3\pi}{2}\,-\,\frac{\pi}{3}\;=\;4\sqrt{3}\,-\,\frac{11\pi}{6}

 
Cool solution. I thought that was a fun little problem.
 
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