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# Help! Anyone!

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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
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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

CPhill  Feb 6, 2018
edited by CPhill  Feb 6, 2018
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Thank you, CPhill!

tertre  Feb 6, 2018
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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