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