rangpur 12th bdmo p5

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barnik
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rangpur 12th bdmo p5

Unread post by barnik » Sat Dec 20, 2014 10:00 am

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ABE is a right angled isosceles triangle. If AE is parallel to CG and BE is parallel DF, AB square = AD square + BC square and AC = 1cm, then what is the ratio between areas of triangle CDH quadrilateral EFGH.

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Raiyan Jamil
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Re: rangpur 12th bdmo p5

Unread post by Raiyan Jamil » Sat Dec 20, 2014 1:35 pm

Very easy.The ratio is 1:1. :D
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Ragib Farhat Hasan
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Re: rangpur 12th bdmo p5

Unread post by Ragib Farhat Hasan » Sun Dec 21, 2014 4:51 pm

How did you get the answer 1:1?

barnik
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Re: rangpur 12th bdmo p5

Unread post by barnik » Thu Dec 25, 2014 8:56 am

how? :shock: :shock: :shock:

tanmoy
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Re: rangpur 12th bdmo p5

Unread post by tanmoy » Thu Dec 25, 2014 11:30 am

$EF\parallel GH,FH\parallel EG.\therefore EFHG$ is a parallelogram.
Now,$\frac{BA}{BC}=\frac{BE}{BG}\Leftrightarrow BC=BG$.$\therefore GE=1$.$\therefore (EFHG)=BD.$
Now,given that,$AB^{2}=AD^{2}+BC^{2}$.
$\therefore (AC+CD+BD)^{2}=(AC+CD)^{2}+(BD+DC)^{2}$.
$\therefore 2BD=CD^{2},BD=\frac{CD^{2}}{2}$.
Now,$(CDH)=\frac{CD^{2}}{2}$.$\therefore$ $\frac{\Delta CDH}{\square EFHG}=1$ :)
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