https://latex.artofproblemsolving.com/e/d/3/ed32b01b9329769b3c0a5200881c125398c4a2f3.png\(In~ the ~diagram,~ angle ~A~ is~ right, ~E ~is ~the ~midpoint ~of ~AB, ~D ~is ~the ~midpoint ~of ~AC, AB = 16 ~and~ AC = 12. ~What ~is ~the ~area ~of ~AEFGD, ~where ~EF ~and ~DG ~are ~perpendicular ~to ~BC?\)
+128570 Since AC = 12.....then AD = 6 = CD
And since AB = 16, then AE = 8 = BE
And since triangle DAE is right.....then DE =sqrt [ AD^2 + AE^2 ] = sqrt [ 6^2 + 8^2 ] = sqrt (100) =10 = GF
And ABC is a 12 - 16 - 20 right triangle
And CG + FB = BC - GF = 20 - 10 = 10
Let FB = x so CG = 10 - x
So
CD^2 - CG^2 = DG^2
And
BE^2 - FB^2 = FE^2
And DG = FE so
6^2 - (10 - x)^2 = 8^2 - x^2 simplify
36 - x^2 + 20x -100 = 64 - x^2
-64 + 20x = 64
20x = 128
x = 128 / 20 = 6.4
So FB = 6.4
And
FE = sqrt ( BE^2 - FB^2) = sqrt ( 8^2 - 6.4^2) = 4.8
So area of right triangle DAE = (1/2)(AD)(AE) = (1/2)(6)(8) = 24 (1)
And the area of DEFG = (DE)(FE) = 10 * 4.8 = 48 (2)
So.....[ AEFGD ] = (1) + (2) = 72 units^2
