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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 64cm2,then the area of triangle aef (in cm2) is ???

matsunnymat  Aug 25, 2015

Best Answer 

 #5
avatar+78754 
+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
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4+0 Answers

 #2
avatar+78754 
+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
 #3
avatar+91051 
0

Nice pic Chris ;)

Melody  Aug 25, 2015
 #4
avatar+78754 
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
avatar+78754 
+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

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