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# S96-04

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Joined: 02 Sep 2009
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16 Sep 2014, 00:50
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Difficulty:

85% (hard)

Question Stats:

61% (02:42) correct 39% (02:30) wrong based on 127 sessions

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In a rectangular coordinate system, point $$A$$ has coordinates $$(d, d)$$, where $$d \gt 0$$. Point $$A$$ and the origin form the endpoints of a diameter of circle $$C$$. What fraction of the area of circle $$C$$ lies within the first quadrant?

A. $$\frac{\pi}{\pi + \sqrt{2}}$$
B. $$\frac{\pi}{\pi + 1}$$
C. $$\sqrt{\frac{2}{\pi}}$$
D. $$\frac{\pi + 2}{2 \pi}$$
E. $$\frac{2 \pi}{2 \pi + 1}$$

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16 Sep 2014, 00:50
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Official Solution:

In a rectangular coordinate system, point $$A$$ has coordinates $$(d, d)$$, where $$d \gt 0$$. Point $$A$$ and the origin form the endpoints of a diameter of circle $$C$$. What fraction of the area of circle $$C$$ lies within the first quadrant?

A. $$\frac{\pi}{\pi + \sqrt{2}}$$
B. $$\frac{\pi}{\pi + 1}$$
C. $$\sqrt{\frac{2}{\pi}}$$
D. $$\frac{\pi + 2}{2 \pi}$$
E. $$\frac{2 \pi}{2 \pi + 1}$$

First, draw the coordinate plane and add in point $$A$$, which has coordinates $$(d,d)$$. Since $$d$$ is greater than 0, this point lies in the first quadrant, as shown:

Notice that the origin, the point $$(0, d)$$, the pint $$(d, 0)$$, and point $$A$$ $$(d,d)$$ form the corners of a square:

Now, we draw the line segment between $$(0,0)$$ and $$(d,d)$$. This is the diameter of a circle, which we also draw:

Notice that the answer does not depend on $$d$$. All that matters is that the circle contains a square. The fraction of the circle's area in the first quadrant will be the same, no matter what. Thus, we can drop $$d$$ as a variable and create a new variable $$r$$ for the radius of the circle. It will be easier to compute areas in terms of $$r$$ (which will also cancel out in the end).

The area of the circle is just $$\pi r^2$$.

The fraction we are looking for is this:

So we need the shaded area of the circle, which consists of the square () and 2 of the four "leaves of the table" ().

The square's area can be found this way:

$$\text{Area} = (\sqrt{2}r)^2 = 2 r^2$$

Now the area of the 4 leaves () is the area of the circle () minus the area of the square (). So the area of the 4 leaves is $$\pi r^2 - 2r^2 = (\pi - 2)r^2$$.

This means that the area of 2 leaves is $$\frac{\pi - 2}{2} r^2$$.

The whole shaded area is thus $$2 r^2 + \leftarrow( \frac{\pi - 2}{2} \rightarrow)r^2 = \leftarrow( 2 + \frac{\pi}{2} - \frac{2}{2} \rightarrow)r^2 = \leftarrow( 1 + \frac{\pi}{2} \rightarrow)r^2$$.

Finally, the desired fraction is $$\frac{\leftarrow( 1 + \frac{\pi}{2} \rightarrow)r^2}{\pi r^2} = \frac{\leftarrow( 1 + \frac{\pi}{2} \rightarrow)X^2}{\pi X^2} = \frac{2 + \pi}{2 \pi}$$.

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26 Nov 2015, 09:18
I think this the explanation isn't clear enough, please elaborate.
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23 Dec 2015, 03:26
the part(s) of the circle that does not lie within the first quadrant are segments of the circle. from the figure it is evident that the central angle of this segment is (pi/2).
area of a segment of a circle = $$(r^2)/2 * (Central angle - sin(Central angle))$$
in this case area of segment = $$(d/2)^2/2*(pi/2 - 1)$$
for two segments, area = $$(d/2)^2*(pi/2 - 1)$$ -------- (1)
area of circle =$$pi * (d/2)^2$$ --------------------------- (2)

so fraction in first quadrant = (2) - (1)

and for our answer = ((2) - (1))/(2)

which will yield (D)
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17 Oct 2016, 18:00
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Don't you think that this problem is too time-consuming for the GMAT? It's doable, though.

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09 Jan 2017, 03:33
I think this is a high-quality question and I agree with explanation.
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16 Feb 2017, 07:08
This is a great question for the GMAT.

Property to be learnt:
Side of Inscribed square = √2 * r ; where r = radius of the circle.
Area of Inscribed square = 2 * r^2 ; where r = radius of the circle.
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09 Apr 2017, 15:27
Bunuel wrote:
In a rectangular coordinate system, point $$A$$ has coordinates $$(d, d)$$, where $$d \gt 0$$. Point $$A$$ and the origin form the endpoints of a diameter of circle $$C$$. What fraction of the area of circle $$C$$ lies within the first quadrant?

A. $$\frac{\pi}{\pi + \sqrt{2}}$$
B. $$\frac{\pi}{\pi + 1}$$
C. $$\sqrt{\frac{2}{\pi}}$$
D. $$\frac{\pi + 2}{2 \pi}$$
E. $$\frac{2 \pi}{2 \pi + 1}$$

I created the same diagram that Bunuel created and it seems like the area out side the first quadrant = 2 parts
and area inside the 1st quadrant = 10 parts
hence answer should be close to 10/12 or 5/6

only answer choice D is close.
Attachments

circle.png [ 16.05 KiB | Viewed 28210 times ]

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22 Jul 2017, 10:27
High Quality Question...It needs lot of practice of similar questions before to do it in simulated time.
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03 Aug 2017, 12:59
Is it reasonable to know that the area of the circle less the area of the square is 2 times the area outside of Quadrant 1? ((Circle - Square) / 2 + Square ) / Circle) is our answer. Seems faster than solving individual secions as it allows you to eliminate most of the variables at the end.

d^2 + d^2 = (Diameter of Circle)^2; sqrt(2d^2) = diameter, radius is sqrt(2d^2)/2

Area of circle is π(sqrt(2d^2)/2)^2 = π(2d^2/4) or πd^2/2;
Area of square is d^2
Area of the circular edges is πd^2/2 - d^2 = d^2(π/2-1)
1/2 of the circular edges are outside Q1 so d^2(π/2-1)/2

Then a matter of adding that area to the square and dividing by the circle

D^2 / 2 cancels out which we needed to happen from the answer choices and we are left with (π+2)/2π
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08 Nov 2017, 20:16
1
2 formula to keep in mind if not derive the sides and use them for square:

Area of square inscribed = 2 X ( Radius)^2
Side of square inscribed in a circle = \sqrt{2} X Radius

Image attached shows the explanation.

All the best,
Nisha

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fullsizeoutput_88.jpeg [ 1.54 MiB | Viewed 18114 times ]

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25 Nov 2017, 14:03
The easiest way : since 4 parts outside the square are equal,
Find area of semi circle + area of Right angled Triangle
This will exclude the 2 parts outside the 1st quadrant
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13 Mar 2018, 03:02
it appears that r in the answer choice is defined as both the diameter and radius.
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10 Oct 2018, 18:35
Agree with Amit. The required area in Q-1 is that of one semi circle + a 90:45:45 right angle with sides d. The hypotenuse of triangle is the diameter (2^0.5)d is the diameter. Rest can be solved easily.
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12 Nov 2018, 00:35
Cant i simply subtract the area of the 2 semicircles forming outside the 1st quadrant from the total area of the circle ?
In that case , wont the radius of the 2 semi circles outside the Q1 be equal to r only ?

Re: S96-04 &nbs [#permalink] 12 Nov 2018, 00:35
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# S96-04

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