Three of the four vertices of a rectangle are (5,11), (6,11) and (16,-2). What is the area of the intersection of this rectangular region and the region inside the graph of the equation (x - 5)^2 + (y + 2)^2 = 16? Express your answer as a common fraction in terms of pi.
+133 The three vertices you specified don't make a rectangle.
Assuming you mean (6,-2):
We can split this region into a circular segment an isosceles triangle. The area of the triangle is 2. I don't have the time to write my process down, but it involved coord bashing and heron's formula.
The angle of the circular sector is arccos(√15/4), so the degrees value is arccos(√15/4)*180/pi. Divide that by 360, we get that the area is (arccos(√15/4)*180/pi)/360 ths of the whole circle's area, or 16 pi * (arccos(√15/4)*180/pi)/360. Subtract 2, we get \(8 \cdot \text{arccos}\left(\frac{\sqrt15}{4}\right) -2 \text{, or } 4 \pi - 8 \cdot \text{arctan}(\sqrt{15}) - 2\)
It's pretty likely that I'm wrong, but perhaps you can identify my mistake by looking over the steps.
https://www.desmos.com/calculator/uys2utn9vh
