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What is the area of the region enclosed by lines y=x, x=−y,
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m22 q20What is the area of the region enclosed by lines \(y=x\), \(x=y\), and the upper crescent of the circle \(y^2+x^2=4\) ? A. \(\frac{\pi}{4}\) B. \(\frac{\pi}{2}\) C. \(\frac{3\pi}{4}\) D. \(\pi\) E. \(4\pi\)
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Originally posted by Rock750 on 06 Apr 2013, 10:17.
Last edited by Bunuel on 18 Feb 2018, 10:46, edited 2 times in total.
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What is the area of the region enclosed by lines y=x, x=−y,
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07 Apr 2013, 03:38
m22 q20What is the area of the region enclosed by lines \(y=x\), \(x=y\), and the upper crescent of the circle \(y^2+x^2=4\) ?A. \(\frac{\pi}{4}\) B. \(\frac{\pi}{2}\) C. \(\frac{3\pi}{4}\) D. \(\pi\) E. \(4\pi\) The circle represented by the equation \(x^2+y^2 = 4\) is centered at the origin and has the radius of \(r=\sqrt{4}=2\). Look at the diagram below: We need to find the area of the upper crescent, so the area of the yellow sector. Since the central angle of this sector is 90 degrees then its area would be 1/4 of that of the circle (since circle is 360 degrees). The area of the circle is \({\pi}{r^2}=4\pi\), 1/4 of this value is \(\pi\). Answer: D. Attachment:
m2220.png [ 16.59 KiB  Viewed 6882 times ]
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Re: What is the area of the region enclosed by lines y=x, x=−y,
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06 Apr 2013, 10:58
Rock750 wrote: What is the area of the region enclosed by lines y=x, x=−y, and the upper crescent of the circle y^2+x^2=4 ?
A \(Pi/4\)
B \(Pi/2\)
C \(3Pi/4\)
D \(Pi\)
E \(4Pi\) \(y^2+x^2=4\) is a circle with its center in the origin (0,0) The lines \(y=x\) and \(y=x\) intersect in (0,0) and form an angle of 90° between them. The area of the circle is \(r^2PI=4PI\), now we are looking for "area of the region enclosed by lines y=x, x=−y, and the upper crescent of the circle y^2+x^2=4", which can be found through this equation \(4PI : 360=x : 90\) (Tot area : Tot angle = x : angBetweenLines) D \(x=PI\)
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Re: What is the area of the region enclosed by lines y=x, x=−y,
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06 Apr 2013, 19:56
The lines y=x, x=−y intersecting at (0,0) and they are perpendicular to each other. The circle y^2+x^2=4 is centered at (0,0) and has a radius of 2. These two lines are dividing the circle y^2+x^2=4 into four equal segments. Area of each segment = (PI*r^2) / 4 = (PI*2^2) / 4 = PI
Answer is D.



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Re: What is the area of the region enclosed by lines y=x, x=−y,
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08 Apr 2013, 11:45
Hi Bunuel, in the above problem, i understand that the circle is centered at the origin and has radius 2 but lines y=x, and x = 1 represents the half of the circle so the area enclosed should be half of the total area, I mean 4pi/2. i know, i am missing something, please clarify! and How do I know that question asks the area of upper 1/4 of the area or How do i determine the 90 degree portion??



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Re: What is the area of the region enclosed by lines y=x, x=−y,
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08 Apr 2013, 12:01
kamalahmmad1 Quote: How do i determine the 90 degree portion Two lines are perpendicular if their slopes are ve reciprocal of each other. E.g. line 1 : y = mx and line 2: y = (1/m)x then line 1 and line two are perpendicular Here y = x and y = x meet the above stated criteria for perpendicular lines //kudos please, if the above explanation is good.
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Re: What is the area of the region enclosed by lines y=x, x=−y,
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09 Apr 2013, 02:46
kamalahmmad1 wrote: Hi Bunuel, in the above problem, i understand that the circle is centered at the origin and has radius 2 but lines y=x, and x = 1 represents the half of the circle so the area enclosed should be half of the total area, I mean 4pi/2. i know, i am missing something, please clarify! and How do I know that question asks the area of upper 1/4 of the area or How do i determine the 90 degree portion?? The two lines are y=x and y=x (x=y), not x = 1. Both lines are shown on the diagram in my post. Lines y=x and y=x make 90 degrees.
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Re: What is the area of the region enclosed by lines y=x, x=−y,
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11 Aug 2013, 09:18
To find the Area of a sector
Arc angle Area of sector ___________ = ______________ 360 Area of circle
90 Area of sector ___ = _____________ 360 4 Pi
Area of sector = 4 pi x 90/ 360 = pi



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Re: What is the area of the region enclosed by lines y=x, x=−y,
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13 Aug 2013, 14:26
Rock750 wrote: m22 q20
What is the area of the region enclosed by lines \(y=x\), \(x=y\), and the upper crescent of the circle \(y^2+x^2=4\) ?
A. \(\frac{\pi}{4}\) B. \(\frac{\pi}{2}\) C. \(\frac{3\pi}{4}\) D. \(\pi\) E. \(4\pi\) ... My solution(with detail explanation): This diagram is indispensable:
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Re: What is the area of the region enclosed by lines y=x, x=−y,
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19 Dec 2014, 03:53
All the 3 parts i.e x=y; x=y; and x^2+y^2=4 makes one forth of the circular region x^2+y^2=4 area of which is 4pi so the required area is pi. hope this one makes it clear.



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Re: What is the area of the region enclosed by lines y=x, x=−y,
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10 Oct 2015, 01:36
sajib2126 wrote: What is the area of the region enclosed by lines y=x, x=−y, and the upper crescent of the circle y^2+x^2=4 ? a.π/4 b. π/2 c. π/4 d. π e.4π Question corrected! Circle is y^2+x^2=4 has radius 2. Area of circle = pi * 4 y=x is the line that passes through 1 and 3 quadrants which has 45 degree to xaxis. y=x is the line that passes through 2 and 4 quadrants which has 45 degree to xaxis. The two lines, divides the circle into 4 parts. Combining all the three, (question to find the area of the upper crescent) Area = 4 pi/4 = pi



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Re: What is the area of the region enclosed by lines y=x, x=−y,
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10 Oct 2015, 01:42
Why or How does Circle is y^2+x^2=4 has radius 2 ?



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Re: What is the area of the region enclosed by lines y=x, x=−y,
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10 Oct 2015, 01:58
sajib2126 wrote: Why or How does Circle is y^2+x^2=4 has radius 2 ? In the equation x^2+y^2=4 all the points (2,2)(2,2)(0,2)(2,0) are on the circle. And these have the center as (0,0) Hence the radius is 2. Another explanation:Equation Of A Circle (x  a)^2 + (y  b)^2 = r^2 where a is the x coordinate of the centre of the circle b is the y coordinate of the centre of the circle r is the radius of the circle Comparing the above equation with the equation (x0)^2+(y0)^2=4 r^2 = 4 r=2 center is (0,0)



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What is the area of the region enclosed by lines y=x, x=−y,
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19 Feb 2018, 14:36
sajib2126 wrote: Why or How does Circle is y^2+x^2=4 has radius 2 ? hi the area of the circle is represented by the equation, x^2 + y^2 = 4, means that the area of the circle is x^2 + y^2 = 4 now, as 2 sides are equal to each other, it can be said that area is (x^2 + y^2) OR 4 since area is 4, the radius is 2 I don't know whether this reasoning is okay, but I have seen the scenario this way Expert's reply on this issue is highly desired thanks ???



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Re: What is the area of the region enclosed by lines y=x, x=−y,
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20 Feb 2018, 19:33
Bunuel wrote: m22 q20What is the area of the region enclosed by lines \(y=x\), \(x=y\), and the upper crescent of the circle \(y^2+x^2=4\) ?A. \(\frac{\pi}{4}\) B. \(\frac{\pi}{2}\) C. \(\frac{3\pi}{4}\) D. \(\pi\) E. \(4\pi\) The circle represented by the equation \(x^2+y^2 = 4\) is centered at the origin and has the radius of \(r=\sqrt{4}=2\). Look at the diagram below: Attachment: m2220.png We need to find the area of the upper crescent, so the area of the yellow sector. Since the central angle of this sector is 90 degrees then its area would be 1/4 of that of the circle (since circle is 360 degrees). The area of the circle is \({\pi}{r^2}=4\pi\), 1/4 of this value is \(\pi\). Answer: D. Hi Bunuel, I was able to solve this question but it took me a while to visualise the diagram. So just asking by curiosity, can we expect this type of questions in GMAT? As always appreciate your efforts in advance! Thanks
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Re: What is the area of the region enclosed by lines y=x, x=−y,
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20 Feb 2018, 19:47
ujjwal80 wrote: Bunuel wrote: m22 q20What is the area of the region enclosed by lines \(y=x\), \(x=y\), and the upper crescent of the circle \(y^2+x^2=4\) ?A. \(\frac{\pi}{4}\) B. \(\frac{\pi}{2}\) C. \(\frac{3\pi}{4}\) D. \(\pi\) E. \(4\pi\) The circle represented by the equation \(x^2+y^2 = 4\) is centered at the origin and has the radius of \(r=\sqrt{4}=2\). Look at the diagram below: Attachment: m2220.png We need to find the area of the upper crescent, so the area of the yellow sector. Since the central angle of this sector is 90 degrees then its area would be 1/4 of that of the circle (since circle is 360 degrees). The area of the circle is \({\pi}{r^2}=4\pi\), 1/4 of this value is \(\pi\). Answer: D. Hi Bunuel, I was able to solve this question but it took me a while to visualise the diagram. So just asking by curiosity, can we expect this type of questions in GMAT? As always appreciate your efforts in advance! Thanks Yes, I think it's a perfectly valid question. Notice that the average time it took users to answer it correctly is just 1:07 minutes.
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What is the area of the region enclosed by lines y=x, x=−y,
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24 Feb 2018, 06:19
ujjwal80 wrote: Bunuel wrote: m22 q20What is the area of the region enclosed by lines \(y=x\), \(x=y\), and the upper crescent of the circle \(y^2+x^2=4\) ?A. \(\frac{\pi}{4}\) B. \(\frac{\pi}{2}\) C. \(\frac{3\pi}{4}\) D. \(\pi\) E. \(4\pi\) The circle represented by the equation \(x^2+y^2 = 4\) is centered at the origin and has the radius of \(r=\sqrt{4}=2\). Look at the diagram below: Attachment: m2220.png We need to find the area of the upper crescent, so the area of the yellow sector. Since the central angle of this sector is 90 degrees then its area would be 1/4 of that of the circle (since circle is 360 degrees). The area of the circle is \({\pi}{r^2}=4\pi\), 1/4 of this value is \(\pi\). Answer: D. Hi Bunuel, I was able to solve this question but it took me a while to visualise the diagram. So just asking by curiosity, can we expect this type of questions in GMAT? As always appreciate your efforts in advance! Thanks hi as far as time is concerned, you can see the problem this way (y =x), means slope is 1 and the line passes through the origin (y = x), means slope is 1, and the line passes through the origin now it is worth noticing that a slope of 1 or 1 creates an angle of 45 degree, so 2 slopes jointly cover a 90 degree also, when 2 lines are perpendicular to each other, their slopes are negative reciprocal to each other here 1 is negative reciprocal to 1, so the slopes are perpendicular to each other, creating an angle 90 degree thus, 2 slopes together add to 90 degrees, which is 1/4 of pi (2)^2 , as the radius is 2 we get from (x^ + Y^ = 4) OR you can simply draw the 2 slopes to see the area covered by the upper crescent hope this helps and is clear! thanks and cheers!



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Re: What is the area of the region enclosed by lines y=x, x=−y,
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24 Feb 2018, 19:01
gmatcracker2018 wrote: ujjwal80 wrote: Bunuel wrote: m22 q20What is the area of the region enclosed by lines \(y=x\), \(x=y\), and the upper crescent of the circle \(y^2+x^2=4\) ?A. \(\frac{\pi}{4}\) B. \(\frac{\pi}{2}\) C. \(\frac{3\pi}{4}\) D. \(\pi\) E. \(4\pi\) The circle represented by the equation \(x^2+y^2 = 4\) is centered at the origin and has the radius of \(r=\sqrt{4}=2\). Look at the diagram below: Attachment: m2220.png We need to find the area of the upper crescent, so the area of the yellow sector. Since the central angle of this sector is 90 degrees then its area would be 1/4 of that of the circle (since circle is 360 degrees). The area of the circle is \({\pi}{r^2}=4\pi\), 1/4 of this value is \(\pi\). Answer: D. Hi Bunuel, I was able to solve this question but it took me a while to visualise the diagram. So just asking by curiosity, can we expect this type of questions in GMAT? As always appreciate your efforts in advance! Thanks hi as far as time is concerned, you can see the problem this way (y =x), means slope is 1 and the line passes through the origin (y = x), means slope is 1, and the line passes through the origin now it is worth noticing that a slope of 1 or 1 creates an angle of 45 degree, so 2 slopes jointly cover a 90 degree also, when 2 lines are perpendicular to each other, their slopes are negative reciprocal to each other here 1 is negative reciprocal to 1, so the slopes are perpendicular to each other, creating an angle 90 degree thus, 2 slopes together add to 90 degrees, which is 1/4 of pi (2)^2 , as the radius is 2 we get from (x^ + Y^ = 4) OR you can simply draw the 2 slopes to see the area covered by the upper crescent hope this helps and is clear! thanks and cheers! Thanks, gmatcracker2018, I will keep your explanation in mind next time I see such question.
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