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Re: In the figure above, E is the center of the circle [#permalink]
3
this question was so what time consuming but easy one

as triangle ACD has one side as diameter of circle so angle C=90

from that diameter \(AD^2=(1/2)^2 + (1/2)^2 \)
AD=(1/√2)
radius=(1/2√2)
so each side of square is = (1/√2)

and

as BG=4AE=√2
Draw a line from B which passes through center of circle and reaches point R and BR=(1/√2)

now triangle BGR is right triangle
\(BG^2 = GR^2 + BR^2 \)
GR=√(3/2)
GF=GR-FR
=\sqrt{(3/2)}-(1/2√2)
= (2√3-1)/2√2
multiply √2 to numerator and denaminator
= (2√6-√2)/4
so option D is correct

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Re: In the figure above, E is the center of the circle [#permalink]
2
Hi guys,

I don´t understand why we can assume B is forming a right triangle with the base of the square. Technically, with no more information given, B could not be forming the 90º triangle with the base, since it could be slightly away from the centre of the square. hence we could not determine the length of GF. I do understand the logic of the problem, but I would like to understand what can we assume and what we can´t. Honestly, the fact that option E is on the table is confusing.

gajala wrote:
this question was so what time consuming but easy one

as triangle ACD has one side as diameter of circle so angle C=90

from that diameter \(AD^2=(1/2)^2 + (1/2)^2 \)
AD=(1/√2)
radius=(1/2√2)
so each side of square is = (1/√2)

and

as BG=4AE=√2
Draw a line from B which passes through center of circle and reaches point R and BR=(1/√2)

now triangle BGR is right triangle
\(BG^2 = GR^2 + BR^2 \)
GR=√(3/2)
GF=GR-FR
=\sqrt{(3/2)}-(1/2√2)
= (2√3-1)/2√2
multiply √2 to numerator and denaminator
= (2√6-√2)/4
so option D is correct

"if you like the explanation please press kudos please"
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Re: In the figure above, E is the center of the circle [#permalink]
A vertical line dropped from point B will only pass through the center C of the Circle if B is the midpoint of the side of the square, but in the problem it is nowhere mentioned that B is the midpoint of the side.
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Re: In the figure above, E is the center of the circle [#permalink]
MarieteC wrote:
Hi guys,

I don´t understand why we can assume B is forming a right triangle with the base of the square. Technically, with no more information given, B could not be forming the 90º triangle with the base, since it could be slightly away from the centre of the square. hence we could not determine the length of GF. I do understand the logic of the problem, but I would like to understand what can we assume and what we can´t. Honestly, the fact that option E is on the table is confusing.

gajala wrote:
this question was so what time consuming but easy one

as triangle ACD has one side as diameter of circle so angle C=90

from that diameter \(AD^2=(1/2)^2 + (1/2)^2 \)
AD=(1/√2)
radius=(1/2√2)
so each side of square is = (1/√2)


Bump
and

as BG=4AE=√2
Draw a line from B which passes through center of circle and reaches point R and BR=(1/√2)

now triangle BGR is right triangle
\(BG^2 = GR^2 + BR^2 \)
GR=√(3/2)
GF=GR-FR
=\sqrt{(3/2)}-(1/2√2)
= (2√3-1)/2√2
multiply √2 to numerator and denaminator
= (2√6-√2)/4
so option D is correct

"if you like the explanation please press kudos please"
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Re: In the figure above, E is the center of the circle [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: In the figure above, E is the center of the circle [#permalink]
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