My Math Forum what is the length

 Geometry Geometry Math Forum

November 8th, 2018, 02:25 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.
Attached Images
 20181106_231032-5.jpg (33.8 KB, 7 views)

Last edited by skipjack; November 8th, 2018 at 07:46 AM.

November 8th, 2018, 04:45 AM   #2
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,590
Thanks: 1434

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.

Attached Images
 Clipboard01.jpg (9.3 KB, 24 views)

November 8th, 2018, 05:45 AM   #3
Newbie

Joined: Oct 2018
From: Turkey

Posts: 23
Thanks: 0

Quote:
 Originally Posted by romsek 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.
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 07:47 AM.

 November 8th, 2018, 05:58 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 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, 06:12 AM   #5
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,590
Thanks: 1434

Quote:
 Originally Posted by skipjack 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.
Is that what those squiggles mean?

Do you think the square at A and C means those two angles are equal?

Last edited by skipjack; November 8th, 2018 at 07:48 AM.

November 8th, 2018, 06:55 AM   #6
Newbie

Joined: Oct 2018
From: Turkey

Posts: 23
Thanks: 0

Quote:
 Originally Posted by romsek Is that what those squiggles mean? Do you think the square at A and C means those two angles are equal?
Square at A and C means they are right angles.

Last edited by skipjack; November 8th, 2018 at 07:48 AM.

 November 8th, 2018, 07:04 AM #7 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 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, 07:20 AM #8 Senior Member     Joined: Feb 2010 Posts: 714 Thanks: 151 Connect B with D and let $\displaystyle BD=d$. Also, let $\displaystyle BA=BE=y$. Then from right triangle $\displaystyle ABD$ we get $\displaystyle x^2=d^2-y^2$. From right triangle $\displaystyle BDC$ we get $\displaystyle BC^2=d^2-7^2$. From right triangle $\displaystyle BEC$ we get $\displaystyle BC^2=y^2-5^2$. So, $\displaystyle d^2-49=y^2-25$ and $\displaystyle d^2-y^2=49-25=24$. Thus, $\displaystyle x^2=24$ and $\displaystyle x=2 \sqrt{6}$. Thanks from greg1313
 November 8th, 2018, 07:35 AM #9 Senior Member     Joined: Feb 2010 Posts: 714 Thanks: 151 Here is a picture. pic1.jpg Thanks from Denis
November 8th, 2018, 09:15 AM   #10
Newbie

Joined: Oct 2018
From: Turkey

Posts: 23
Thanks: 0

Quote:
 Originally Posted by mrtwhs Connect B with D and let $\displaystyle BD=d$. Also, let $\displaystyle BA=BE=y$. Then from right triangle $\displaystyle ABD$ we get $\displaystyle x^2=d^2-y^2$. From right triangle $\displaystyle BDC$ we get $\displaystyle BC^2=d^2-7^2$. From right triangle $\displaystyle BEC$ we get $\displaystyle BC^2=y^2-5^2$. So, $\displaystyle d^2-49=y^2-25$ and $\displaystyle d^2-y^2=49-25=24$. Thus, $\displaystyle x^2=24$ and $\displaystyle x=2 \sqrt{6}$.
Thanks a lot

 Tags length

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post mms Math 10 July 31st, 2015 07:11 AM joshbeldon Calculus 3 February 26th, 2015 11:58 AM amin7905 Algebra 4 July 19th, 2012 01:08 PM kfarnan Calculus 1 November 26th, 2011 11:56 AM future_prodegy Calculus 1 April 12th, 2009 06:24 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top