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In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1), and (-1, 1). If a point falls into the square region, what is the probability that the ordinates of the point (x, y) satisfy that \(x^2+y^2>1\)?
(A) \(1-\frac{\pi}{4}\)
(B) \(\frac{\pi}{2}\)
(C) \(4-\pi\)
(D) \(2-\pi\)
(E) \(\pi-2\)
(A) \(1-\frac{\pi}{4}\)
(B) \(\frac{\pi}{2}\)
(C) \(4-\pi\)
(D) \(2-\pi\)
(E) \(\pi-2\)
21 Dec 2009, 04:40
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delta09
First note that the square we have is centered at the origin, has the length of the sides equal to 2 and the area equal to 4.
\(x^2+y^2=1\) is an equation of a circle also centered at the origin, with radius 1 and the \(area=\pi{r^2}=\pi\).
We are told that the point is IN the square and want to calculate the probability that it's outside the circle (\(x^2+y^2>1\) means that the point is outside the given circle):
P = (Favorable outcome)/(Total number of possible outcomes).
Favorable outcome is the area between the circle and the square=\(4-\pi\)
Total number of possible outcomes is the area of the square (as given that the point is in the square) =\(4\)
\(P=\frac{4-\pi}{4}=1-\frac{\pi}{4}\)
Answer: A.
Hope it's clear.
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General Discussion
21 Dec 2009, 04:37
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delta09
\(x^2 + y^2 = R^2\) is the equation of a circle with centre (0,0) and radius R
If we draw the square and the circle, then we will see that the circle is inscribed in the square i.e. the diameter of the circle is equal to the length of the side of square.
Area of the circle= pi (1)^2=pi
Area of the square=(2)^2 = 4
The proabibility that the point lies within the square and outside the circle is
(Area of the sqaure - area of the circle)/area of the circle
= 4-pi/4
=1 - pi/4
Plz. let me know if the OA is A.
Thanks
