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What is the area of the region enclosed by lines y=x, x=−y,

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What is the area of the region enclosed by lines y=x, x=−y,  [#permalink]

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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. \(4\pi\)

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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.
Edited the question.
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What is the area of the region enclosed by lines y=x, x=−y,  [#permalink]

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New post 07 Apr 2013, 03:38
8
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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. \(4\pi\)

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:

Image

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:
m22-20.png
m22-20.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,  [#permalink]

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New post 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,  [#permalink]

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New post 06 Apr 2013, 19:56
1
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,  [#permalink]

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New post 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,  [#permalink]

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New post 08 Apr 2013, 12:01
2
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,  [#permalink]

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New post 09 Apr 2013, 02:46
1
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,  [#permalink]

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New post 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,  [#permalink]

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New post 13 Aug 2013, 14:26
3
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:
Attachments

line and circle.png
line and circle.png [ 42.47 KiB | Viewed 6077 times ]


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Re: What is the area of the region enclosed by lines y=x, x=−y,  [#permalink]

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New post 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,  [#permalink]

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New post 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 x-axis.
y=-x is the line that passes through 2 and 4 quadrants which has 45 degree to x-axis.

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,  [#permalink]

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New post 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,  [#permalink]

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New post 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 co-ordinate of the centre of the circle
b is the y co-ordinate of the centre of the circle
r is the radius of the circle

Comparing the above equation with the equation (x-0)^2+(y-0)^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,  [#permalink]

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New post 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
???
:roll:
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Re: What is the area of the region enclosed by lines y=x, x=−y,  [#permalink]

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New post 20 Feb 2018, 19:33
Bunuel 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. \(4\pi\)

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:
m22-20.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,  [#permalink]

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New post 20 Feb 2018, 19:47
1
ujjwal80 wrote:
Bunuel 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. \(4\pi\)

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:
m22-20.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,  [#permalink]

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New post 24 Feb 2018, 06:19
ujjwal80 wrote:
Bunuel 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. \(4\pi\)

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:
m22-20.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!
:cool:
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Re: What is the area of the region enclosed by lines y=x, x=−y,  [#permalink]

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New post 24 Feb 2018, 19:01
gmatcracker2018 wrote:
ujjwal80 wrote:
Bunuel 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. \(4\pi\)

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:
m22-20.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!
:cool:


Thanks, gmatcracker2018, I will keep your explanation in mind next time I see such question.
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