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# In the figure below, if isosceles right angle PQR has an

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In the figure below, if isosceles right angle PQR has an [#permalink]  19 Jun 2004, 08:56
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In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!

A. Pi
B. 2Pi
C. 2. sqrt2. pi
D. 4 pi
E. 8 pi
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Last edited by mirhaque on 19 Jun 2004, 09:45, edited 1 time in total.
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Since R is center, let RP = RQ = x
Area of PRQ = 1/2*RQ)RP =x^2/2
Area given = 4
Hence x^2/2 = 4 =>x = 2*sqrt2!!
For this Isoceles triangles if we drop a perpendicular from R (RY) to base PQ, it will bisect PQ at Y.
From pythagoras theorem, PQ^2 = PR^2 + RQ^2 =>PQ = 4 => QY =2!!

Hence RY^2 = RQ^2 - QY^2 = {2*sqrt2}^2 -{2}^2 = 4
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Re: PS: GEOMETRY [#permalink]  20 Jun 2004, 02:32
mirhaque wrote:
In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!

A. Pi
B. 2Pi
C. 2. sqrt2. pi
D. 4 pi
E. 8 pi

S(triangle) = 4 => Radius of the circle = 2 (since R*2*R/2 = area = 4). => S(shaded region) = S(circle)/4 = 2^2*pi/4 = pi.
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Remember, if you get booged down with calculations you are going to be screwed, and this question tries to lead you down that path.
The triangle has an area of 4, and the shaded portion has an area obviuosly less than that. Only one answer choice fits that criteria.
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Maneesh wrote:
Remember, if you get booged down with calculations you are going to be screwed, and this question tries to lead you down that path.
The triangle has an area of 4, and the shaded portion has an area obviuosly less than that. Only one answer choice fits that criteria.

No, Maneesh, you are wrong!

2*sqrt(pi) ~~ 2*1.8 = 3.6 < 4!
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Re: PS: GEOMETRY [#permalink]  24 Jun 2004, 08:55
Emmanuel wrote:
mirhaque wrote:
In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!

A. Pi
B. 2Pi
C. 2. sqrt2. pi
D. 4 pi
E. 8 pi

S(triangle) = 4 => Radius of the circle = 2 (since R*2*R/2 = area = 4). => S(shaded region) = S(circle)/4 = 2^2*pi/4 = pi.

I completely understood CBRF3's explanation, but your condensed version is not as clear. Just following up in case you got some great shortcut.
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Re: PS: GEOMETRY [#permalink]  24 Jun 2004, 09:17
lastochka wrote:
Emmanuel wrote:
mirhaque wrote:
In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!

A. Pi
B. 2Pi
C. 2. sqrt2. pi
D. 4 pi
E. 8 pi

S(triangle) = 4 => Radius of the circle = 2 (since R*2*R/2 = area = 4). => S(shaded region) = S(circle)/4 = 2^2*pi/4 = pi.

I completely understood CBRF3's explanation, but your condensed version is not as clear. Just following up in case you got some great shortcut.

S(circle) is the area of the circle.

We find area of triangle = 4 => we find all sides of this triangle (2*sqrt(2),2*sqrt(2), 4). Then we find its height (from the vertex which is opposite to hipotenuse), which is equal to 2(because hipotenuse = 4, area = 4 => hipotenuse*height/2 = area = 4). This height is the same as the radius of the circle! Then we find area of the circle, given its radius: pi*r^2 = 4*PI. Then we find shaded area, which is one-fourth of the circle area = PI.
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Re: PS: GEOMETRY [#permalink]  24 Jun 2004, 11:08
Emmanuel wrote:
lastochka wrote:
Emmanuel wrote:
mirhaque wrote:
In the figure below, if isosceles right angle PQR has an area of 4, what is the area of the shaded portion of the fgure? Assume R is the center. PLEASE EXPLAIN!

A. Pi
B. 2Pi
C. 2. sqrt2. pi
D. 4 pi
E. 8 pi

S(triangle) = 4 => Radius of the circle = 2 (since R*2*R/2 = area = 4). => S(shaded region) = S(circle)/4 = 2^2*pi/4 = pi.

I completely understood CBRF3's explanation, but your condensed version is not as clear. Just following up in case you got some great shortcut.

S(circle) is the area of the circle.

We find area of triangle = 4 => we find all sides of this triangle (2*sqrt(2),2*sqrt(2), 4). Then we find its height (from the vertex which is opposite to hipotenuse), which is equal to 2(because hipotenuse = 4, area = 4 => hipotenuse*height/2 = area = 4). This height is the same as the radius of the circle! Then we find area of the circle, given its radius: pi*r^2 = 4*PI. Then we find shaded area, which is one-fourth of the circle area = PI.

in other words, it's the same throught process as presented by CBRF3. thanks for elaboration
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Maneesh wrote:
Remember, if you get booged down with calculations you are going to be screwed, and this question tries to lead you down that path.
The triangle has an area of 4, and the shaded portion has an area obviuosly less than that. Only one answer choice fits that criteria.

good work Maneesh.

Emmanuel, the third option is 2.sqrt2.pi which is > 4

- ash
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ash
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ashkg wrote:

Emmanuel, the third option is 2.sqrt2.pi which is > 4

- ash

No, ashkg, 2*sqrt(PI) < 4 because 4*PI < 16, because PI = 3.14 < 4.
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Emmanuel wrote:
ashkg wrote:

Emmanuel, the third option is 2.sqrt2.pi which is > 4

- ash

No, ashkg, 2*sqrt(PI) < 4 because 4*PI < 16, because PI = 3.14 < 4.

Emmanuel, what you have written is correct, but what is given is 2.srqt2.pi not 2*sqrt(PI). Maneesh, thanks. You are thinking outside the box. It took me several minutes to find the answer using traditional math.
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