abc ia triangle .d is mid point of ab. e is the midpoint of db .fis the mid point of bc.if the area of triangle abc is 64cm^{2},then the area of triangle aef (in cm^{2}) is ???

matsunnymat
Aug 25, 2015

#5**+10 **

It appears that this is always true, even if we have an obtuse triangle.....see the pic......

The reason for this is that ΔAEF will always have (1/2) the height of ΔABC, and it's base will always be (3/4) of ΔABC.

So Area of ΔABC = (1/2)bh = (1/2)(8)(16) = 64 cm^2

And Area of ΔAEF = (1/2)([3/4]*8)([1/2]*16) = (1/2)(6)(8) = 24 cm^2

CPhill
Aug 25, 2015

#2**+10 **

Here is a picture of what you describe :

Δ ABC has a base = 16 and a height = 8 ....so the area is (16)(8)/2 = 64cm^2

D is the midpoint of AB, E is the midpoint of DB, and F is the midpoint of BC. And since D has the same x coordinate as C, then the x coordinate of F will have the same x coordinate as the midpoint of DB. And the y coordinate of F will be the same as the y coordinate of the midpoint of DC.

Thus, the height of Δ AEF = 4 and the base = 12. So the area of Δ AEF is just (12)(4)/2 = 48/2 = 24 cm^2

CPhill
Aug 25, 2015

#4**0 **

Thanks, Melody......I'm not sure that this answer is unique....for instance, if triangle ABC had a different shape [but the same area as above], would triangle AEF still have the same area???......I don't really know.....maybe someone else could prove/disprove it.....!!!!

CPhill
Aug 25, 2015

#5**+10 **

Best Answer

It appears that this is always true, even if we have an obtuse triangle.....see the pic......

The reason for this is that ΔAEF will always have (1/2) the height of ΔABC, and it's base will always be (3/4) of ΔABC.

So Area of ΔABC = (1/2)bh = (1/2)(8)(16) = 64 cm^2

And Area of ΔAEF = (1/2)([3/4]*8)([1/2]*16) = (1/2)(6)(8) = 24 cm^2

CPhill
Aug 25, 2015