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## Dhaka Secondary 2010/7

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**don't post problems (by starting a topic)**in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can 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.

### Dhaka Secondary 2010/7

In the figure above $AD = 4, AB = 3$ and $CD = 9$. What is the area of triangle $\triangle AEC$?

### Re: Dhaka Secondary 2010/7

It's not that hard; come on guys!

"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please

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.- leonardo shawon
**Posts:**169**Joined:**Sat Jan 01, 2011 4:59 pm**Location:**Dhaka

### Re: Dhaka Secondary 2010/7

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

Ibtehaz Shawon

BRAC University.

BRAC University.

*long way to go .....*- Tahmid Hasan
**Posts:**665**Joined:**Thu Dec 09, 2010 5:34 pm**Location:**Khulna,Bangladesh.

### Re: Dhaka Secondary 2010/7

$$\frac{9}{2}$$ i guess.i just used similar triangle properties and Pythagoras

বড় ভালবাসি তোমায়,মা

### Re: Dhaka Secondary 2010/7

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, $(ACE) = (ACD) - (CDE) = \frac {4.9}2 - \frac {3.9}2 = \frac 92$.

$\angle ABC = \angle DCB$.

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.****"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes**