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Dhaka Secondary 2010/7

Posted: Fri Jan 21, 2011 7:06 pm
by BdMO
div geo.png
div geo.png (6.95KiB)Viewed 9265 times
In the figure above $AD = 4, AB = 3$ and $CD = 9$. What is the area of triangle $\triangle AEC$?

Re: Dhaka Secondary 2010/7

Posted: Tue Feb 01, 2011 11:40 pm
by Moon
It's not that hard; come on guys!

Re: Dhaka Secondary 2010/7

Posted: Tue Feb 01, 2011 11:43 pm
by leonardo shawon
I'm working on it. We just need to find the AE?!"

Re: Dhaka Secondary 2010/7

Posted: Tue Feb 01, 2011 11:46 pm
by Tahmid Hasan
$$\frac{9}{2}$$ i guess.i just used similar triangle properties and Pythagoras :D

Re: Dhaka Secondary 2010/7

Posted: Sun Sep 02, 2012 4:57 pm
by Labib
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, $(ACE) = (ACD) - (CDE) = \frac {4.9}2 - \frac {3.9}2 = \frac 92$.