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

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Math Expert
Joined: 02 Sep 2009
Posts: 54410
Square ABCD has side length 2. A semicircle with diameter AB is constr  [#permalink]

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21 Mar 2019, 05:00
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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 [ 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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21 Mar 2019, 09:39
1
Bunuel wrote:

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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