Calculate area of a crescent

Question

My math teacher gave me this question as a challenge, but I can't seem to solve it. If anyone could kindly assist me in how to go about solving it, and providing a set of detailed steps required to solve it, it would be more appreciated. Below I have provided the following link to the question for which I wish to answer. The question:

Answer 1

There may be a cleverer trick I'm not seeing, but the standard way to solve this problem is with integration. Note that we can overlay the picture on the cartesian plane with the functions

1. $f(x) = \sqrt{25-x^2}$ (the upper half of the circle)
2. $g(x) = -\sqrt{25-x^2}$ (the lower half of the circle)
3. $h(x)=\sqrt{100-(x-5)^2}-5$ (the big circle whose centre is at the bottom-right of the picture)

Link to graph if it helps

Recall that we can get the area under a function $\phi(x)$ and above $\psi(x)$ from $x=\alpha$ to $x=\beta$ by means of the integral $$\int_\alpha^\beta \phi(x) - \psi(x).$$ We start our integration at the left end of the circle, where $f(x)$ intersects $g(x)$. This is at $x=-5$. The function $f$ is always directly above the crescent, so this is the upper function, but we need to find out where $g(x)$ meets $h(x)$, so we can switch the lower function at some point. With some calculation, we find that $g(x) = h(x)$ at $x = -5(\sqrt 7 + 1)/4$. Lets call this value $a$ for short. So part of the area is described by the integral $$A_1 = \int_{-5}^a f(x) - g(x).$$ The second integral will start at $x=a$, but to find its upper bound, we need to figure out the where $f(x)$ and $h(x)$ intersect. This time we find that $f(x) = h(x)$ intersect at $x = 5(\sqrt7 - 1)/4$. Call this value $b$, and the part of the area we're missing is $$A_2 = \int_a^b f(x) - h(x).$$ The total area of the crescent is $A_1+A_2$.