In the diagram, AB, BC, CD, DE, EF, FG, GH, and HK all have length 4, and all angles are right angles, with the exception of the angles at D and F. Find the area of the polygon.
\(\phantom{what is x^2}\)
+128407 Draw DF and CG
Really....the only trick here is to find DF = sqrt [ 4^2 + 4^2 ] = sqrt (32) = 4sqrt 2 = CG
Area of BHKA = [ BA * (BC + DF + GH) ] = [ 4 * ( 4 + 4sqrt 2 + 4) ] = [ 4 ( 8 + 4sqrt 2 ) =
32 + 16sqrt 2
Area of rectangle DFGC = FG * DF = 4 * 4sqrt 2 = 16sqrt 2
Area of right triangle DFE = (1/2) * DE * FE = (1/2) * 4^2 = 8
Area of DEFGC = 16sqrt 2 - 8
Area of polygon =
32 +16sqrt 2 + 16sqrt 2 - 8 =
24 + 32sqrt 2
