find area of shaded region; two circles

[galactus]

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.

 

[soroban]

Hello, galactus!

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

              * * *
          *           *
        *               *
       *                 *

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

Draw \(\displaystyle PQ\). \(\displaystyle \Q\,=\,4\)

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


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

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

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


Therefore, the shaded region has area: \(\displaystyle \:4\sqrt{3}\,-\,\frac{3\pi}{2}\,-\,\frac{\pi}{3}\;=\;4\sqrt{3}\,-\,\frac{11\pi}{6}\)

 

[galactus]

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