An easy geometry problem. Show your work.
The problem will be attached.

Question

An easy geometry problem. Show your work.
The problem will be attached.

anonymous · Answer

attachment

anonymous · Answer

This problem was generated in on my site http://saab.org

ganeshie8 · Answer

I've just figured out the solution xD I'll leave a hint maybe : 
All equilateral triangles are similar

mathmath333 · Answer

it will be $20\sqrt 3$

as $A(\triangle ABC)/A(\triangle MNP)=4/1$

ganeshie8 · Answer

BINGO!
small correction : 
ratio of areas = 4
that means ratio of sides has to be 2 right ? so we should get  $$\dfrac{AB}{MN}=2 \implies MN = 20$$

anonymous · Answer

eh i dont know i'm not convinced 
its sounds like
|dw:1423220299880:dw|

anonymous · Answer

i only know that the big triangle side is 40 , and area btw are equivalent that tells me it can be any triangle inside could be the one we want :|

anonymous · Answer

@ganeshie8  and @mathmath333  are right

mathmath333 · Answer

yes its 40!!

anonymous · Answer

oh wait also the small triangle are the same to the bounded i missed that xD

anonymous · Answer

I just did it and got a) 10

anonymous · Answer

Pull the triangle downwards.
|dw:1423220496831:dw|
Let the area of each part is x cm^2.
Now, both the triangles are similar and you can take the ratio of the areas and you get the sides.