1. ## Find the area

Good Afternoon!!!

Please, can you help me with it?

Where AB = 5, AD = 3, CE = 1. The area of the triangle BCF is?

I started making some analogy between similar triangles:

First: between ABF and CEF

But Im not sure...

What can I do now?!

2. Originally Posted by PxEckx
Good Afternoon!!!

Please, can you help me with it?

Where AB = 5, AD = 3, CE = 1. The area of the triangle BCF is?

I started making some analogy between similar triangles:

First: between ABF and CEF

But Im not sure...

What can I do now?!
Yes, you can continue along that route if you like.
Triangles ABF and CEF are similar. ABF is CEF magnified.

The base of ABF is 5 and the base of CEF is 1.
This means that the perpendicular height of CEF is one-fifth of the perpendicular height of ABF,
since the linear measurements of ABF are 5 times those of CEF.

Hence you must divide AD or BC into 6 equal subdivisions.
Then you have the base and perpendicular height of CEF,
so you can find it's area.

Can you finish ?

OOps!!! sorry, that's ECF not BCF.

However, you then only need subtract the area of ECF from ECB which is easy.

3. Ahhhh! Now I undertood!

I made the 'divison' as you said (but of the BCF!) , and I fond the height!

$\displaystyle h=\displaystyle{\frac{5}{6}}$

Then I just calculate the area!

$\displaystyle \displaystyle{\frac{BC.h}{2} = \displaystyle{\frac{5}{4}$

Thanx!!!!

4. Well done!

Quite clever...
You found 5/6 of EC.