+0  
 
-6
175
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avatar+-280 

In square $ABCD$ with sides of length 4 cm, $N$ is the midpoint of side $BC$ and $M$ is the midpoint of side $CD$. What is the area of triangle $AMN$, 

 Apr 1, 2020
 #1
avatar+202 
-1

How are you still awake? Imma go to bed soon. Sorry.

 Apr 1, 2020
 #3
avatar+112419 
+2

Note  that  the  area  of the square  = 4^2   = 16 cm^2

 

Triangles   ADM  and ABN  are  both  right.....their  combined  areas  = (2)(4)  =  8  cm^2

 

Triangle MNC is also right and has an area  =  (1/2)(2)(2)   =  2 cm^2

 

So....the area of triangle  AMN =   [16  -  8   -  2 ] cm^2    =   6  cm^2

 

 

cool cool cool

 Apr 1, 2020

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