AB.docx- Confusing figure
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Confusing geometry
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Raiyan Jamil
- Posts:138
- Joined:Fri Mar 29, 2013 3:49 pm
In the attachment figure below , the rectangle ABCD , along the line BE the triangle ABE is folded and the point A falls on CD at A' . BA' = 10 , BC =6 , DE = a/b and a , b are co-prime , a+b = ? and what is the value of a and b ?
Last edited by Raiyan Jamil on Sat Feb 08, 2014 8:19 pm, edited 1 time in total.
A smile is the best way to get through a tough situation, even if it's a fake smile.
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Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Re: Confusing geometry
$11$,$8$,$3$
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.
Re: Confusing geometry
@Thanic Nur Samin Please post full solution or hints with the answer.
@Raiyan I do not know about others but I am having difficulty viewing the image properly.
If you it is possible, could you upload the image in a better format like .png or .jpeg ?
You can use Geogebra or Paint to get this job done.
@Raiyan I do not know about others but I am having difficulty viewing the image properly.
If you it is possible, could you upload the image in a better format like .png or .jpeg ?
You can use Geogebra or Paint to get this job done.
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
Re: Confusing geometry
$AB=B{A}',AE=E{A}'$.Applying Pythagorus's theorem,we get,${A}'C=8.\therefore D{A}'=2$.Suppose,$AE=x$.$\therefore DE=6-x$
$\therefore (6-x)^{2}+4=x^{2}\Rightarrow x=\frac{10}{3}.\therefore 6-x=\frac{8}{3}.
\therefore DE=\frac{8}{3}.\therefore a+b=11,a=8,b=3$.
$\therefore (6-x)^{2}+4=x^{2}\Rightarrow x=\frac{10}{3}.\therefore 6-x=\frac{8}{3}.
\therefore DE=\frac{8}{3}.\therefore a+b=11,a=8,b=3$.
"Questions we can't answer are far better than answers we can't question"
Re: Confusing geometry
AB=BAAE=EA .Applying Pythagorus's theorem,we get,AC=8DA=2.Suppose,AE=x.DE=6−x
(6−x)2+4=x2x=3106−x=38DE=38a+b=11a=8b=3.
(6−x)2+4=x2x=3106−x=38DE=38a+b=11a=8b=3.