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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
 #1
avatar+92808 
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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
 #2
avatar+3462 
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Thank you, CPhill! smiley

tertre  Feb 6, 2018
 #3
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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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