Dhaka Secondary 2010/7

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Secondary 2010/7

Unread post by BdMO » Fri Jan 21, 2011 7:06 pm

div geo.png
div geo.png (6.95 KiB) Viewed 2114 times
In the figure above $AD = 4, AB = 3$ and $CD = 9$. What is the area of triangle $\triangle AEC$?

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Moon
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Re: Dhaka Secondary 2010/7

Unread post by Moon » Tue Feb 01, 2011 11:40 pm

It's not that hard; come on guys!
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

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leonardo shawon
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Re: Dhaka Secondary 2010/7

Unread post by leonardo shawon » Tue Feb 01, 2011 11:43 pm

I'm working on it. We just need to find the AE?!"
Ibtehaz Shawon
BRAC University.

long way to go .....

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Tahmid Hasan
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Re: Dhaka Secondary 2010/7

Unread post by Tahmid Hasan » Tue Feb 01, 2011 11:46 pm

$$\frac{9}{2}$$ i guess.i just used similar triangle properties and Pythagoras :D
বড় ভালবাসি তোমায়,মা

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Labib
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Re: Dhaka Secondary 2010/7

Unread post by Labib » Sun Sep 02, 2012 4:57 pm

As $AB || CD$ and $\angle DAB=\angle ADC = 90^{\circ}$,
$\angle ABC = \angle DCB$.
Thus, $\Delta ABE \sim \Delta CDE$.
As $AE:DE = AB:CD = 3:9 = 1:3$,
$DE = 3$.
Therefore, $(ACE) = (ACD) - (CDE) = \frac {4.9}2 - \frac {3.9}2 = \frac 92$.
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.


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