Thread: 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?!

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.

Ahhhh! Now I undertood!

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



Then I just calculate the area!



Thanx!!!!

Well done!

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