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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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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

30% (01:48) correct 70% (03:10) wrong based on 10 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) \(5 - \sqrt{5}\)

Attachment:
daef6f3ebe3e1fc6cfacd8611db1d6d449c8d2d4.png
daef6f3ebe3e1fc6cfacd8611db1d6d449c8d2d4.png [ 6.42 KiB | Viewed 307 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
IMO D
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Re: Square ABCD has side length 2. A semicircle with diameter AB is constr   [#permalink] 21 Mar 2019, 09:39
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