Three of the four vertices of a rectangle are (5,11), (16,11) 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 =9? Express your answer as a common fraction in terms of .
+36916 The center of the circle is on the 4rth corner of the rectangle (5, -2) and has radius of 3
1/4 of the circle is with in the rectangle
area of circle = pi r^2
1/4 of this = 1/4 pi (3)^2 = 9/4 pi = 7.068 sq units
+228 9pi /4
First, make you can make a good guess of what the missing point is by looking at the graph. (5,-2)
https://www.desmos.com/calculator
The you can see that a radius created here and the intersecting area is essentially an area of a sector, which (1/2) * (r^2) * the angle of the sector.
