November 8th, 2018, 03:25 AM  #1 
Newbie Joined: Oct 2018 From: Turkey Posts: 23 Thanks: 0  what is the length
What is x in the attached? All I can think of is to draw a line from A to E, but that doesn't seem to help much either. Answer is 2 * sqrt 6, but I do not know how they obtained this. Last edited by skipjack; November 8th, 2018 at 08:46 AM. 
November 8th, 2018, 05:45 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,266 Thanks: 1198 
First off ABCD is a rectangle even if you choose not to draw it that way. Second with the information show the problem is indeterminate as the following picture illustrates. 
November 8th, 2018, 06:45 AM  #3 
Newbie Joined: Oct 2018 From: Turkey Posts: 23 Thanks: 0  Did you see that AB and BE are equal to each other? And how can you say ABCD is a rectangle? What makes you think the other two corners are right angles? I didn't catch that part.
Last edited by skipjack; November 8th, 2018 at 08:47 AM. 
November 8th, 2018, 06:58 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,110 Thanks: 1909 
I disagree with the above. I doubt that it's intended that there are right angles at A and C, and even if there are, it doesn't seem to follow that ABCD is a rectangle. Also, the rectangle diagrams don't support BA = BE. However, a plausible diagram would be a good idea. 
November 8th, 2018, 07:12 AM  #5  
Senior Member Joined: Sep 2015 From: USA Posts: 2,266 Thanks: 1198  Quote:
Do you think the square at A and C means those two angles are equal? Last edited by skipjack; November 8th, 2018 at 08:48 AM.  
November 8th, 2018, 07:55 AM  #6 
Newbie Joined: Oct 2018 From: Turkey Posts: 23 Thanks: 0  Square at A and C means they are right angles.
Last edited by skipjack; November 8th, 2018 at 08:48 AM. 
November 8th, 2018, 08:04 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,110 Thanks: 1909 
I think ketanco is mistaken. My analysis indicates that the angles at A and C are equal, but are not right angles. Also, a correct diagram is quite surprising and very different in appearance from the deliberately misdrawn diagram provided. However, I've not yet found a proof that the diagram I have in mind is unique. 
November 8th, 2018, 08:20 AM  #8 
Senior Member Joined: Feb 2010 Posts: 701 Thanks: 136 
Connect B with D and let $\displaystyle BD=d$. Also, let $\displaystyle BA=BE=y$. Then from right triangle $\displaystyle ABD$ we get $\displaystyle x^2=d^2y^2$. From right triangle $\displaystyle BDC$ we get $\displaystyle BC^2=d^27^2$. From right triangle $\displaystyle BEC$ we get $\displaystyle BC^2=y^25^2$. So, $\displaystyle d^249=y^225$ and $\displaystyle d^2y^2=4925=24$. Thus, $\displaystyle x^2=24$ and $\displaystyle x=2 \sqrt{6}$. 
November 8th, 2018, 10:15 AM  #10  
Newbie Joined: Oct 2018 From: Turkey Posts: 23 Thanks: 0  Quote:
 

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