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Square ABCD has side length 2. A semicircle with diameter AB is constr

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Square ABCD has side length 2. A semicircle with diameter AB is constr  [#permalink]

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New post 21 Mar 2019, 05:00
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Question Stats:

29% (02:01) correct 71% (03:29) wrong based on 21 sessions

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Square ABCD has side length 2. A semicircle with diameter AB is constructed inside the square, and the tangent to the semicircle from C intersects side AD at E. What is the length of CE ?


(A) \(\frac{2 + \sqrt{5}}{2}\)

(B) \(\sqrt{5}\)

(C) \(\sqrt{6}\)

(D) 5/2

(E) \(5 - \sqrt{5}\)

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daef6f3ebe3e1fc6cfacd8611db1d6d449c8d2d4.png [ 6.42 KiB | Viewed 861 times ]

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Re: Square ABCD has side length 2. A semicircle with diameter AB is constr  [#permalink]

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New post 21 Mar 2019, 09:39
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Bunuel wrote:
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Square ABCD has side length 2. A semicircle with diameter AB is constructed inside the square, and the tangent to the semicircle from C intersects side AD at E. What is the length of CE ?


(A) \(\frac{2 + \sqrt{5}}{2}\)

(B) \(\sqrt{5}\)

(C) \(\sqrt{6}\)

(D) 5/2

(E) \(5 - \sqrt{5}\)

Attachment:
daef6f3ebe3e1fc6cfacd8611db1d6d449c8d2d4.png


let tangent of the line CE , touch semicircle at F
so side

BC = FC = 2 and AE = EF = x. Thus DE = 2-x.
Pythagorean Theorem on triangle CDE yields

\begin{align*} DE^2 + CD^2 &= CE^2\\ (2-x)^2 + 2^2 &= (2+x)^2\\ x^2 - 4x + 8 &= x^2 + 4x + 4\\ x &= \frac{1}{2}\end{align*}

CE= FC+ x= 5/2
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Re: Square ABCD has side length 2. A semicircle with diameter AB is constr  [#permalink]

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New post 27 May 2019, 17:55
Can someone please give another way to approach this problem?

Kind regards!
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Square ABCD has side length 2. A semicircle with diameter AB is constr  [#permalink]

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New post 27 May 2019, 18:20
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jfranciscocuencag wrote:
Can someone please give another way to approach this problem?

Kind regards!


jfranciscocuencag

The one posted above is the best way, check the theory and if you come across tangents GMAT will definitely be testing this theory:-

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Capture 1.PNG [ 25.48 KiB | Viewed 414 times ]


Attachment:
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Capture 2.PNG [ 29.21 KiB | Viewed 414 times ]

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Square ABCD has side length 2. A semicircle with diameter AB is constr  [#permalink]

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New post 27 May 2019, 23:21
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jfranciscocuencag wrote:
Can someone please give another way to approach this problem?

Kind regards!


Hi jfranciscocuencag,

Perhaps this will help:

Theorem:
This theorem states that if from one external point, two tangents are drawn to a circle then they have equal tangent segments.

So point c is the external point from where two tangents are drawn,one at point B and the other at point F,(F is the Point of intersection of CE with the semicircle).Hence these two tangent segments will have equal length 2.

Similarly consider E as the external point and two tangents are drawn to points A and F, again these two tangent's will be equal.

Hope now it's clear?
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Square ABCD has side length 2. A semicircle with diameter AB is constr   [#permalink] 27 May 2019, 23:21
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