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24 Dec 2008, 04:31
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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?
* \(\frac{\pi}{4}\)
* \(\frac{\pi}{2}\)
* \(\pi\)
* \(2\pi\)
* \(3\pi\)
* \(\frac{\pi}{4}\)
* \(\frac{\pi}{2}\)
* \(\pi\)
* \(2\pi\)
* \(3\pi\)
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24 Dec 2008, 05:08
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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
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
02 Dec 2009, 13:22
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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?
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?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
