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# m14 q28

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05 Jan 2010, 22:39
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If 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?

(A) $$\frac{\pi}{4}$$
(B) $$\frac{\pi}{2}$$
(C) $$\pi$$
(D) $$2\pi$$
(E) $$3\pi$$

Source: GMAT Club Tests - hardest GMAT questions

The area of the circle is $$\pi 2^2 = 4\pi$$ . The area of the sector = $$\text{(the area of the circle)}*\frac{90}{360} = \pi$$ .

Shouldn't the answer be B, since the top part of the sector is only 45degrees. The whole thing is 90 degrees, but half of it is at the bottom of the xy plane.

REVISED VERSION OF THIS QUESTION IS HERE: m14-q28-88810.html#p1237818
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06 Jan 2010, 05:11
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Expert's post
11MBA wrote:
If 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?

(C) 2008 GMAT Club - m14#28

* $$\frac{\pi}{4}$$
* $$\frac{\pi}{2}$$
* $$\pi$$
* $$2\pi$$
* $$3\pi$$

The area of the circle is $$\pi 2^2 = 4\pi$$ . The area of the sector = $$\text{(the area of the circle)}*\frac{90}{360} = \pi$$ .

Shouldn't the answer be B, since the top part of the sector is only 45degrees. The whole thing is 90 degrees, but half of it is at the bottom of the xy plane.

$$x^2 + y^2 = 4$$ is an equation of a circle centered at the origin and the radius $$\sqrt{4}=2$$.

Graph of the function $$y = |x|$$ is given below:
Attachment:

graph_modulus.png [ 7.61 KiB | Viewed 8485 times ]

Top part of the sector would be a sector with 90 degrees angle, and its area would be 1/4 of that of the circle (circle 360 degrees).

Area of the circle =$${\pi}{r^2}=4\pi$$, 1/4 of this value = $$\pi$$.
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16 Jun 2011, 10:42
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Area of sector = Θ/360°*πr²
Now we know that r= 2 and Θ= 90°
So, area comes to be π.
Attachments

Blank.png [ 196.1 KiB | Viewed 7100 times ]

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22 Jan 2013, 00:28
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Why don't we extend both the lines and account for the 45 degree angle formed in the quadrants III & IV ? I am sorry if its a silly question.
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06 Jan 2010, 08:07
My bad. I graphed abso(y)=x, and that's why i only had half of the graph be above the xy plane. Sorry.
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14 Jun 2010, 07:59
I know these are supposed to be the toughest challenges, but is this something which is included in the scope of the GMAT? I thought coordinate plane on GMAT was restricted to straight line equations only....

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14 Jun 2010, 22:14
From x^2 + y^2 = r^2 => radius = 2
Area of the circle = [img]\pi[/img]r^2
The line |x| has a reflection about the origin at [img]90^{\circ}[/img]
Area of sector is 1/4 x area of circle = [img]\pi[/img]4/4 = [img]\pi[/img]
Therefore ans is C
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13 Aug 2010, 01:48
Its C.

The sector makes an angle of 90 degrees at the center of the circle. So area of the sector = area of the circle/4 = pi
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16 Jun 2011, 09:59
I did not know the equation for a circle, so had to spend extra seconds figuring out that x^2 + Y^2 = 4 was indeed the equation of a circle with radius 2. I ended up with C, but probably took 30-60s more than I should have.

Question - do these type of problems represent the problems at the 700-800 level on the GMAT? I thought the equation for a circle was pretty much out of scope for the GMAT. I hear that the GMAT quant section is getting tougher but is this (problems like this) the level where its headed?
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16 Jun 2011, 10:33
ulm wrote:
I saw a lot of questions with circle formulas (both in paper and digital tests) and suppose it's 650 question maximum.

Well, then I'm glad I came across this problem Could you tell me in which paper and digital tests you saw questions with circle formulas? I've pretty much based my Math theory on the Manhattan guides and don't remember seeing the circle formula there (or for that matter OG11).
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16 Jun 2011, 11:15
AriBenCanaan wrote:
ulm wrote:
I saw a lot of questions with circle formulas (both in paper and digital tests) and suppose it's 650 question maximum.

Well, then I'm glad I came across this problem Could you tell me in which paper and digital tests you saw questions with circle formulas? I've pretty much based my Math theory on the Manhattan guides and don't remember seeing the circle formula there (or for that matter OG11).

If we're talking about must-have tests, it definitely was in GMAT Prep.
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17 Jun 2011, 07:43
y = |x| ---- the slope of this line is (+1) if X is positive and (-1) if X is negative...which means the lines make a 45 degree angle with the x axis...and both the lines sweep a total of 90 degrees...

The circle is 360 degrees -- so the area swept is 1/4 of the total area.

(x^2) + (y^2) = (2)^2

Area of the circle = pi*(2^2)

Area of the sector = (1/4)*Area of the circle

Ans: pi (C)
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17 Jun 2011, 11:04
C

the sector has a 90 degree angle..
area of a sector theta(90 degree)/360 x (pi x sq. radius)
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20 Jun 2012, 06:04
Straight C.

As seen in Microstrip's diagram, the area is a quarter of the total area of the circle.
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22 Jan 2013, 06:40
Expert's post
jainpiyushjain wrote:
Why don't we extend both the lines and account for the 45 degree angle formed in the quadrants III & IV ? I am sorry if its a silly question.

Meanwhile go through this post for a solution: m14-q28-88810.html#p670247

Hope it helps.
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28 Jan 2013, 10:53
Bunuel wrote:
11MBA wrote:
If 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?

(C) 2008 GMAT Club - m14#28

* $$\frac{\pi}{4}$$
* $$\frac{\pi}{2}$$
* $$\pi$$
* $$2\pi$$
* $$3\pi$$

The area of the circle is $$\pi 2^2 = 4\pi$$ . The area of the sector = $$\text{(the area of the circle)}*\frac{90}{360} = \pi$$ .

Shouldn't the answer be B, since the top part of the sector is only 45degrees. The whole thing is 90 degrees, but half of it is at the bottom of the xy plane.

$$x^2 + y^2 = 4$$ is an equation of a circle centered at the origin and the radius $$\sqrt{4}=2$$.

Graph of the function $$y = |x|$$ is given below:
Attachment:
graph_modulus.png

Top part of the sector would be a sector with 90 degrees angle, and its area would be 1/4 of that of the circle (circle 360 degrees).

Area of the circle =$${\pi}{r^2}=4\pi$$, 1/4 of this value = $$\pi$$.

How did you find the radius "$$x^2 + y^2 = 4$$ is an equation of a circle centered at the origin and the radius $$\sqrt{4}=2$$." can you please explain?
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28 Jan 2013, 11:00
Expert's post
kuttingchai wrote:
Bunuel wrote:
11MBA wrote:
If 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?

(C) 2008 GMAT Club - m14#28

* $$\frac{\pi}{4}$$
* $$\frac{\pi}{2}$$
* $$\pi$$
* $$2\pi$$
* $$3\pi$$

The area of the circle is $$\pi 2^2 = 4\pi$$ . The area of the sector = $$\text{(the area of the circle)}*\frac{90}{360} = \pi$$ .

Shouldn't the answer be B, since the top part of the sector is only 45degrees. The whole thing is 90 degrees, but half of it is at the bottom of the xy plane.

$$x^2 + y^2 = 4$$ is an equation of a circle centered at the origin and the radius $$\sqrt{4}=2$$.

Graph of the function $$y = |x|$$ is given below:
Attachment:
graph_modulus.png

Top part of the sector would be a sector with 90 degrees angle, and its area would be 1/4 of that of the circle (circle 360 degrees).

Area of the circle =$${\pi}{r^2}=4\pi$$, 1/4 of this value = $$\pi$$.

How did you find the radius "$$x^2 + y^2 = 4$$ is an equation of a circle centered at the origin and the radius $$\sqrt{4}=2$$." can you please explain?

Check here: math-coordinate-geometry-87652.html (Circle on a plane chapter).

Hope it helps.
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20 Jun 2013, 05:26
Expert's post
11MBA wrote:
If 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?

(A) $$\frac{\pi}{4}$$
(B) $$\frac{\pi}{2}$$
(C) $$\pi$$
(D) $$2\pi$$
(E) $$3\pi$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

The area of the circle is $$\pi 2^2 = 4\pi$$ . The area of the sector = $$\text{(the area of the circle)}*\frac{90}{360} = \pi$$ .

Shouldn't the answer be B, since the top part of the sector is only 45degrees. The whole thing is 90 degrees, but half of it is at the bottom of the xy plane.

BELOW IS REVISED VERSION OF THIS QUESTION:

What is the area of the smaller sector enclosed by the graphs $$x^2 + y^2 = 4$$ and $$y = |x|$$?

A. $$\frac{\pi}{4}$$
B. $$\frac{\pi}{2}$$
C. $$\pi$$
D. $$2\pi$$
E. $$3\pi$$

$$x^2 + y^2 = 4$$ is an equation of a circle centered at the origin and with the radius of $$\sqrt{4}=2$$.

The graph of $$y = |x|$$ is given below (blue):
Attachment:

Curves.png [ 13.45 KiB | Viewed 4230 times ]

The smaller sector enclosed by these graphs would be the top part of the circle, so 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 $$\pi$$.

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11 Jun 2014, 06:49
The smaller sector enclosed by these graphs would be the top part of the circle, so 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 $$\pi$$.

Please, could you explain what does the "smaller sector" mean? It's smaller than what?
The greater part of the circle, not enclosed by y=|x|, is considered a sector too?
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11 Jun 2014, 08:34
X^2+y^2= 4 is a equation of circle centered at origin.
this circle has a radius of 2.
y= lxl is a equation of straight line in the first and the second quadrant and makes an angle of 45 degree with y axis in both the quadrant.
So a total of 90 degree. It means that line occupies 1/4 th of a circle
so net area= (pi (2)^2)/4
=pi.
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Re: m14 q28   [#permalink] 11 Jun 2014, 08:34

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# m14 q28

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