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

Question

https://gmatclub.com/forum/download/file.php?id=47026 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} daef6f3ebe3e1fc6cfacd8611db1d6d449c8d2d4.png

Archit3110 · Answer

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

jfranciscocuencag · Answer

Can someone please give another way to approach this problem?

Kind regards!

LeoN88 · Answer

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:- Capture 1.PNG Capture 2.PNG

stne · Answer

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

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

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