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 ketanco November 8th, 2018 02:25 AM

what is the length

1 Attachment(s)
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.

 romsek November 8th, 2018 04:45 AM

1 Attachment(s)
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.

http://mymathforum.com/attachment.ph...1&d=1541684711

 ketanco November 8th, 2018 05:45 AM

Quote:
 Originally Posted by romsek (Post 602139) 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. http://mymathforum.com/attachment.ph...1&d=1541684711
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.

 skipjack November 8th, 2018 05:58 AM

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.

 romsek November 8th, 2018 06:12 AM

Quote:
 Originally Posted by skipjack (Post 602142) 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?

 ketanco November 8th, 2018 06:55 AM

Quote:
 Originally Posted by romsek (Post 602143) 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.

 skipjack November 8th, 2018 07:04 AM

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.

 mrtwhs November 8th, 2018 07:20 AM

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 $\displaystyle x=2 \sqrt{6}$.

 mrtwhs November 8th, 2018 07:35 AM

1 Attachment(s)
Here is a picture.
Attachment 10018

 ketanco November 8th, 2018 09:15 AM

Quote:
 Originally Posted by mrtwhs (Post 602148) 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

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