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# Curves

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Senior Manager
Joined: 05 Oct 2008
Posts: 272
Followers: 3

Kudos [?]: 122 [0], given: 22

Curves [#permalink]  24 Dec 2008, 03:31
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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
Curves $$x^2 + y^2 = 4$$ and $$y = |x|$$ enclose a sector on the top part of XY-plane. What is the area of this sector?

* $$\frac{\pi}{4}$$
* $$\frac{\pi}{2}$$
* $$\pi$$
* $$2\pi$$
* $$3\pi$$
Director
Joined: 01 Apr 2008
Posts: 906
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 18

Kudos [?]: 311 [2] , given: 18

Re: Curves [#permalink]  24 Dec 2008, 04:08
2
KUDOS
x^2 + y^2 = 4 is a circle with center (0,0) and radius 2. Plot it on a paper.
y=x is a line passing through (0,0) at 45degrees in the second quadrant.
y=-x is a line passing through (0,0) at 45 degrees in the first quadrant.

so the angle between two lines is 2(pi/4) = pi/2 = 90 degrees
area of sector = (pi * r^2 ) / 4 = pi
ans. C
Intern
Joined: 31 Oct 2009
Posts: 38
Followers: 1

Kudos [?]: 34 [0], given: 0

Re: Curves [#permalink]  02 Dec 2009, 12:22
I plotted both curves on paper and it was a little easier for me to see that way.

The first curve is a circle about the origin with a radius of 2
The second curve is a "V" shape--two $$45^{\circ}$$ lines going through the top 2 quadrants meeting at the origin

The area of the first curve = $$\pi2^2=4\pi$$
The area of the first curve in the top 2 quadrants is $$\frac{4\pi}{2}=2\pi$$

Each "leg" of the "V" of the second curve clearly bisects one quarter of the circle created by the first curve, so the area is $$\frac{2\pi}{2}=\pi$$

Where does this question come from?
Re: Curves   [#permalink] 02 Dec 2009, 12:22
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