BDMO Divisional_2014

Forum rules
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.
User avatar
Tasnood
Posts: 11
Joined: Tue Jan 06, 2015 1:46 pm

BDMO Divisional_2014

Unread post by Tasnood » Tue Jan 24, 2017 10:42 pm

The area of ABC and OBC triangle is 120 and 24 respectively. BC=16, EF=8. Find out the area of OEAF Quadrilateral.
Attachments
Capture.JPG
Capture.JPG (13.2 KiB) Viewed 301 times

dshasan
Posts: 66
Joined: Fri Aug 14, 2015 6:32 pm
Location: Dhaka,Bangladesh

Re: BDMO Divisional_2014

Unread post by dshasan » Wed Jan 25, 2017 2:20 pm

$EF = \dfrac{1}{2} BC$ so,$EF \parallel BC$

So, $\bigtriangleup AEF = \dfrac{1}{4}\bigtriangleup ABC = 30$

Now, $\bigtriangleup OEF$ is similar to $\bigtriangleup OBC$

So, $\dfrac{\bigtriangleup OEF}{\bigtriangleup OBC} = \dfrac{EF^2}{BC^2} = \dfrac{1}{4}$

$\Rightarrow \bigtriangleup OEF = 6$

SO, area of $OEAF = \bigtriangleup AEF + \bigtriangleup OEF = 30 + 6 = 36$
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

User avatar
Tasnood
Posts: 11
Joined: Tue Jan 06, 2015 1:46 pm

Re: BDMO Divisional_2014

Unread post by Tasnood » Thu Jan 26, 2017 10:08 pm

You can't say EF||BC only to see that EF=BC/2. Because E and F mayn't be the midpoints. Thanks

Post Reply