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# n the diagram, angle A is right...

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https://latex.artofproblemsolving.com/e/d/3/ed32b01b9329769b3c0a5200881c125398c4a2f3.png\(In~ the ~diagram,~ angle ~A~ is~ right, ~E ~is ~the ~midpoint ~of ~AB, ~D ~is ~the ~midpoint ~of ~AC, AB = 16 ~and~ AC = 12. ~What ~is ~the ~area ~of ~AEFGD, ~where ~EF ~and ~DG ~are ~perpendicular ~to ~BC?\)

Dec 12, 2019

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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   Dec 13, 2019