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 \(\pi\).

tertre
Feb 6, 2018

#3**+2 **

The sides of the rectangle are parallel to the axes, so the fourth point must make a vertical line with (5,11) and a horizontal one with (16,-2); this means that the fourth point is (5,-2). The graph of the region inside the equation is a circle with radius 3 and center (5,-2):

Since each angle of a rectangle is \(90^{\circ}\) and the corner coincides with the center of the circle, the rectangle covers exactly a quarter of the circle. The area of the intersection is thus \(\frac{1}{4}r^2\pi=\frac14\cdot3^2\pi=\boxed{\frac94\pi}\) .

azsun
Feb 6, 2018