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In triangle $ABC$, we have that $E$ and $F$ are midpoints of sides $\overline{AC}$ and $\overline{AB}$, respectively. The area of $\triangle ABC$ is 24 square units. How many square units are in the area of $\triangle CEF$?

 Nov 12, 2017
 #1
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Call AB  the height of  ΔABC and AC the base

 

Then  (1/2) AC * AB  =  24 units^2   ⇒  AC * AB  =  48 units^2

 

But   EC = (1/2)AC  is the base of   ΔCEF  and (1/2)AB =  AF  is the height

 

So  area of    Δ CEF  =  (1/2)CE *AF  =  (1/2) [(1/2)AC] * [(1/2)AB]   =  1/8 AC * AB  =

 

(1/8)* 48  =   6 units^2

 

 

cool cool cool

 Nov 12, 2017

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