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In a rectangular coordinate system, point A has coordinates [#permalink]

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20 Aug 2010, 13:30

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In a rectangular coordinate system, point A has coordinates (d, d), where d > 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?

Re: What fraction of the area of circle C lies within ... [#permalink]

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20 Aug 2010, 13:48

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Ok so I am typing this on my mobile with imagination... Imagine a square of side 1 and a circle circumscribed.. Qn is what is the area that is lying on the 2 bulging sides.. Area of square = 1 area of circle = pi/2

The area of 4 bulging sides = (pi/2-1) Fraction of circle area inside 1st Q= [pi/2-1/2(pi/2-1)]/pi/2 = (pi/2-pi/4+1/2)/(pi/2) = (pi/4+1/2)/(pi/2) = (pi+2)/2pi

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Last edited by mainhoon on 20 Aug 2010, 15:05, edited 2 times in total.

In a rectangular coordinate system, point A has coordinates (d, d), where d > 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?

Answer-choises will come later.

Look a the diagram:

Attachment:

CS.jpg [ 22.42 KiB | Viewed 8701 times ]

We have circle with radius=1 and square inscribed in it. Now imagine the base of square to be x-axis and left side of square to be y-axis. We need to find the ratio of area of a circle without the red parts to the area of whole circle.

Area of a circle is \(C=\pi{r^2}=\pi\); Area of a square is half of the product of diagonals, as diagonal equals to \(2r=2\), then \(S=\frac{2^2}{2}=2\);

Area of a circle without the red parts is \(C-\frac{C-S}{2}=\pi-\frac{\pi-2}{2}=\frac{\pi+2}{2}\);

Ratio of the are of this region to area of a circle is \(\frac{\frac{\pi+2}{2}}{\pi}=\frac{\pi+2}{2\pi}\).

Re: What fraction of the area of circle C lies within ... [#permalink]

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23 Oct 2010, 12:33

Let us assume A to be (1,1).Then, distance from origin = sqrt 2 = Diagonal of square = Diameter of circle.

Area of square = 1 sq units and Area of circle = Pi r^2 = Pi (sqrt 2 / 2)^2 = Pi / 2

Area of 4 bulging sides = (Pi/2) - 1 = (Pi - 2) / 2 and Area of 2 bulging side = (Pi - 2) / 4.

Therefore, required fraction = Area of 2 bulging side / Area of circle = [(Pi - 2) / 4] / (Pi/2) = Pi -2 / 2Pi.

But answer posted by Bunuel is : Pi + 2 / 2Pi and I take his explainations as absolute. Since I got a different answer....please explain where I am wrong.

Re: What fraction of the area of circle C lies within ... [#permalink]

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19 Apr 2012, 21:08

mainhoon wrote:

Ok so I am typing this on my mobile with imagination... Imagine a square of side 1 and a circle circumscribed.. Qn is what is the area that is lying on the 2 bulging sides.. Area of square = 1 area of circle = pi/2

The area of 4 bulging sides = (pi/2-1) Fraction of circle area inside 1st Q= [pi/2-1/2(pi/2-1)]/pi/2 = (pi/2-pi/4+1/2)/(pi/2) = (pi/4+1/2)/(pi/2) = (pi+2)/2pi

Posted from my mobile device

Great work!!...Appreciate the stratgy.
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Re: In a rectangular coordinate system, point A has coordinates [#permalink]

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08 Aug 2012, 07:57

Hi, Cant we do this by calculating the area of the sector and then the area of the circle??? Pls help me understand!
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Re: In a rectangular coordinate system, point A has coordinates [#permalink]

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08 Aug 2012, 16:41

es we can do it by taking any value for (d,d) and calculating the area deducting the sector. Take (d,d) as 2,2. the dia meter becomes 2 root 2 and radius root2 so area of circle = 2pie area of 2 sectors= 2*(90/360)*2 pie = pie But have to add another half of square = 1/2 * 4 = 2 So area in 1st quadrant = 2pie-pie+2= pie+2 and area of circle = 2pie So ratio= (2+pie)/2pie
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Re: What fraction of the area of circle C lies within ... [#permalink]

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23 Jan 2013, 09:15

Bunuel wrote:

Financier wrote:

In a rectangular coordinate system, point A has coordinates (d, d), where d > 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?

Answer-choises will come later.

Look a the diagram:

Attachment:

CS.jpg

We have circle with radius=1 and square inscribed in it. Now imagine the base of square to be x-axis and left side of square to be y-axis. We need to find the ratio of area of a circle without the red parts to the area of whole circle.

Area of a circle is \(C=\pi{r^2}=\pi\); Area of a square is half of the product of diagonals, as diagonal equals to \(2r=2\), then \(S=\frac{2^2}{2}=2\);

Area of a circle without the red parts is \(C-\frac{C-S}{2}=\pi-\frac{\pi-2}{2}=\frac{\pi+2}{2}\);

Ratio of the are of this region to area of a circle is \(\frac{\frac{\pi+2}{2}}{\pi}=\frac{\pi+2}{2\pi}\).

Answer: D.

Hope it's clear.

Hi,

Area of a circle without the red parts is \(C-\frac{C-S}{2}=\pi-\frac{\pi-2}{2}=\frac{\pi+2}{2}\);

is already describing the circle's portion, which is available in 1st Q. why we have again check for the ratio b/w this and circle again.I got this doubt since question has asked us to find the fraction of the area of circle C lies within the first quadrant.

Re: What fraction of the area of circle C lies within ... [#permalink]

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23 Jan 2013, 13:19

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FTG wrote:

Bunuel wrote:

Financier wrote:

In a rectangular coordinate system, point A has coordinates (d, d), where d > 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?

Answer-choises will come later.

Look a the diagram:

Attachment:

CS.jpg

We have circle with radius=1 and square inscribed in it. Now imagine the base of square to be x-axis and left side of square to be y-axis. We need to find the ratio of area of a circle without the red parts to the area of whole circle.

Area of a circle is \(C=\pi{r^2}=\pi\); Area of a square is half of the product of diagonals, as diagonal equals to \(2r=2\), then \(S=\frac{2^2}{2}=2\);

Area of a circle without the red parts is \(C-\frac{C-S}{2}=\pi-\frac{\pi-2}{2}=\frac{\pi+2}{2}\);

Ratio of the are of this region to area of a circle is \(\frac{\frac{\pi+2}{2}}{\pi}=\frac{\pi+2}{2\pi}\).

Answer: D.

Hope it's clear.

Hi,

Area of a circle without the red parts is \(C-\frac{C-S}{2}=\pi-\frac{\pi-2}{2}=\frac{\pi+2}{2}\);

is already describing the circle's portion, which is available in 1st Q. why we have again check for the ratio b/w this and circle again.I got this doubt since question has asked us to find the fraction of the area of circle C lies within the first quadrant.

Kindly help me understand ..

The question asks for "fraction" or "proportion" of the circle. (e.g. say 1/2 or 3/5th of circle lies in the first quadrant) Hence you need to take the ratio of Area (in 1st quadrant) to Total area to find the fraction.
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Re: What fraction of the area of circle C lies within ... [#permalink]

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04 Feb 2013, 08:37

Bunuel wrote:

Financier wrote:

In a rectangular coordinate system, point A has coordinates (d, d), where d > 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?

Answer-choises will come later.

Look a the diagram:

Attachment:

CS.jpg

We have circle with radius=1 and square inscribed in it. Now imagine the base of square to be x-axis and left side of square to be y-axis. We need to find the ratio of area of a circle without the red parts to the area of whole circle.

Area of a circle is \(C=\pi{r^2}=\pi\); Area of a square is half of the product of diagonals, as diagonal equals to \(2r=2\), then \(S=\frac{2^2}{2}=2\);

Area of a circle without the red parts is \(C-\frac{C-S}{2}=\pi-\frac{\pi-2}{2}=\frac{\pi+2}{2}\);

Ratio of the are of this region to area of a circle is \(\frac{\frac{\pi+2}{2}}{\pi}=\frac{\pi+2}{2\pi}\).

Answer: D.

Hope it's clear.

Quote:

Area of a square is half of the product of diagonals, as diagonal equals to \(2r=2\), then \(S=\frac{2^2}{2}=2\);

is this because you are taking into account the portion of the square which doesnt touch the red parts of the circle? Since area of Square is A = a^2 ?

In a rectangular coordinate system, point A has coordinates (d, d), where d > 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?

Answer-choises will come later.

Look a the diagram:

Attachment:

CS.jpg

We have circle with radius=1 and square inscribed in it. Now imagine the base of square to be x-axis and left side of square to be y-axis. We need to find the ratio of area of a circle without the red parts to the area of whole circle.

Area of a circle is \(C=\pi{r^2}=\pi\); Area of a square is half of the product of diagonals, as diagonal equals to \(2r=2\), then \(S=\frac{2^2}{2}=2\);

Area of a circle without the red parts is \(C-\frac{C-S}{2}=\pi-\frac{\pi-2}{2}=\frac{\pi+2}{2}\);

Ratio of the are of this region to area of a circle is \(\frac{\frac{\pi+2}{2}}{\pi}=\frac{\pi+2}{2\pi}\).

Answer: D.

Hope it's clear.

Quote:

Area of a square is half of the product of diagonals, as diagonal equals to \(2r=2\), then \(S=\frac{2^2}{2}=2\);

is this because you are taking into account the portion of the square which doesnt touch the red parts of the circle? Since area of Square is A = a^2 ?

If I understand correctly you are asking about the area of a square: \(area_{square}=side^2=\frac{diagonal^2}{2}\). This is a general formula for the area of any square.

Forgot to mention that the area of a rhombus is also equals to half of the product of diagonals: \(area_{rhombus}=\frac{d_1*d_2}{2}\), where \(d_1\) and \(d_2\) are the lengths of the diagonals (or \(bh\), where \(b\) is the length of the base and \(h\) is the altitude).
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Re: In a rectangular coordinate system, point A has coordinates [#permalink]

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22 Aug 2014, 14:54

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Re: In a rectangular coordinate system, point A has coordinates [#permalink]

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Re: In a rectangular coordinate system, point A has coordinates [#permalink]

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19 Aug 2016, 22:59

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A quick look at the answer choices tells you that the ratio does not depend on the value of "d". Hence you can choose "smart numbers" for d - For example 2. Which makes the radius of the circle sqrt(2). The all we need to do is add the area of the semi circle with the area of the triangle to get the area in the first quadrant (refer the diagram by Bunuel).

Re: In a rectangular coordinate system, point A has coordinates [#permalink]

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19 Aug 2016, 23:11

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megha_2709 wrote:

why we are considering square for this question, its nowhere specified in the question. I do not get this at all, please help.

Regards Megha

If you drop perpendiculars to x & y axes from (d,d) you will get (d,0) and (0,d). These two points are "d" units away from the origin as well as from (d,d). Knowing that all sides are of the same length and that the 3 angles are 90 degrees, (we drew perpendiculars to the axes and the axes themselves are at a () degree angle) we can conclude that the figure is a square.

Hope it helps.

gmatclubot

Re: In a rectangular coordinate system, point A has coordinates
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19 Aug 2016, 23:11

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