Area Of A Sickle

We have split this diagram into 4 areas to find the area of the exterior part. Note that the total square area is 9.

The area of A1 is $9-\frac{9\pi}{4}$ , which can be shown by noticing the area of the quadrant $DAB$ is $\frac {9\pi}{4}$ .

Similarly, the area of A2 is $4-\pi$ .

The area of A3 is 1.

The area of A4 can be determined by finding the area of the of the quadrant that defines it's largest arc ( $DK$ ), and subtracting the area of the quadrant defines the smallest arc ( $JK$ ). Thus, A4 is equal to $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$

Thus, the total exterior area is $9 - \frac {9\pi}{4} + 4 - \pi + 1 + \frac {3\pi}{4} = 14 - \frac {5\pi}{2}$ . Thus, the area of the sickle is $9 - (14 - \frac {5\pi}{2}) = \frac{5\pi-10}{2}$ . Thus, $a=5$ , $b=-10$ and $c=2$ , so $a+b+c=-3$ .

We can use the integration rule to systematically find the area $A$ as follow.

$\begin{aligned} A & = \int\limits_{B \to B} f(x) \ dx \\ & = \int\limits_{B \to J} f(x) \ dx \ {\color{#D61F06}- \int\limits_{J \to K} f(x) \ dx} + \int\limits_{K \to D} f(x) \ dx \ {\color{#D61F06}- \int\limits_{D \to B} f(x) \ dx} & \small {\color{#D61F06} - \text{ sign for reverse direction}} \\ & = \pi {\color{#D61F06} - \left(2 - \frac \pi 4 \right)} + \left(6 - \pi \right) {\color{#D61F06} - \left(9 - \frac {9\pi} 4 \right)} \\ & = \frac {5\pi}2 - 5 = \frac {5\pi-10}2 \end{aligned}$

$\implies a+b+c = 5-10+2 = \boxed{-3}$

The desired answer is $5-10+2=\boxed{-3}$

Define a crescent to be a quarter circle minus the triangle formed by its 3 vertices. For example, $[\mbox{Crescent } BD] = [\mbox{Quadrant } ABD]-[\triangle ABD]$

Remove crescent KJ and translate and rotate crescent BJ to crescent KD. Now we can see that the area is made up of crescent KJ and crescent BD. By similarity, $[\mbox{crescent } BD] = 9[\mbox{crescent } KJ]$ . $\mbox{Area}=[\mbox{crescent } BD] + [\mbox{crescent } KJ]$ $=10[\mbox{crescent } KJ]$ $=10[\frac{\pi}{4}-\frac12(1)(1)]$ $=\boxed{\frac{5\pi-10}{2}}$