Answer:
4x+3=3x+9 given B is mid point of AC.
4x-3x=9-3
x=9-3
x=6
Width of a rectangle is half its length:
2w = l
Perimeter is 36
2w + 2l = 36
Plug first equation into second (substitution)
2w + 2(2w) = 36
2w + 4w = 36
6w = 36
w = 6
Plug back into first equation to solve for l
2(6) = 12 = l
Check the picture below.
so if y = 2, then 2y = 4, thus
![\textit{area of a segment of a circle}\\\\A=\cfrac{r^2}{2}\left(\cfrac{\pi \theta }{180}~~ - ~~sin(\theta ) \right)~~\begin{cases}r=radius\\\theta =\stackrel{degrees}{angle}\\[-0.5em]\hrulefill\\r=4\\\theta =60\end{cases}\\\\\\A=\cfrac{4^2}{2}\left(\cfrac{\pi (60) }{180}~~ - ~~sin(60^o ) \right)\implies A=8\left( \cfrac{\pi }{3}~~ - ~~\cfrac{\sqrt{3}}{2} \right)\\\\\\A=\cfrac{8\pi }{3}~~ - ~~4\sqrt{3}\implies A\approx 1.45](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20segment%20of%20a%20circle%7D%5C%5C%5C%5CA%3D%5Ccfrac%7Br%5E2%7D%7B2%7D%5Cleft%28%5Ccfrac%7B%5Cpi%20%5Ctheta%20%7D%7B180%7D~~%20-%20~~sin%28%5Ctheta%20%29%20%20%5Cright%29~~%5Cbegin%7Bcases%7Dr%3Dradius%5C%5C%5Ctheta%20%3D%5Cstackrel%7Bdegrees%7D%7Bangle%7D%5C%5C%5B-0.5em%5D%5Chrulefill%5C%5Cr%3D4%5C%5C%5Ctheta%20%3D60%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5CA%3D%5Ccfrac%7B4%5E2%7D%7B2%7D%5Cleft%28%5Ccfrac%7B%5Cpi%20%2860%29%20%7D%7B180%7D~~%20-%20~~sin%2860%5Eo%20%29%20%20%5Cright%29%5Cimplies%20A%3D8%5Cleft%28%20%5Ccfrac%7B%5Cpi%20%7D%7B3%7D~~%20-%20~~%5Ccfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%20%5Cright%29%5C%5C%5C%5C%5C%5CA%3D%5Ccfrac%7B8%5Cpi%20%7D%7B3%7D~~%20-%20~~4%5Csqrt%7B3%7D%5Cimplies%20A%5Capprox%201.45)
The distance between the two is 22 units.