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rangpur 12th bdmo p5
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
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.
- Raiyan Jamil
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- Joined:Fri Mar 29, 2013 3:49 pm
Re: rangpur 12th bdmo p5
Very easy.The ratio is 1:1.
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Re: rangpur 12th bdmo p5
How did you get the answer 1:1?
Re: rangpur 12th bdmo p5
$EF\parallel GH,FH\parallel EG.\therefore EFHG$ 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,$(CDH)=\frac{CD^{2}}{2}$.$\therefore$ $\frac{\Delta CDH}{\square EFHG}=1$
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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