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+0 Answers


The center of this circle will lie on the remaining vertex.....


The  intersection of these two areas is just 1/4 of the circle's area  =


pi* r^2  / 4  =     pi (3)^2 / 4    =    (9/4) pi  units^2


cool cool cool

CPhill  Feb 6, 2018
edited by CPhill  Feb 6, 2018

Thank you, CPhill! smiley

tertre  Feb 6, 2018

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

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