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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 .

Guest Dec 6, 2018
 #1
avatar+14579 
+2

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

ElectricPavlov  Dec 6, 2018
 #2
avatar+118 
+5

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.

echofire18  Dec 6, 2018

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