+0

# impossible geometry

0
68
1

ABCD is a square with side length 12, BE = BC/2, FC = DC/3, GD = DA/4, AH = AE/3, JE = EF/2.  Find the area of the purple triangle. May 3, 2020

#1
+1

It's  just really computing the areas of 6 right triangles and subtracting these combined areas from the area of the square

First right triangle  ABE

The legs are BE  = 6     AB  =12

Area  = (1/2) prduct of the legs  = (1/2)(6)(12) = 36

We also  need to find AE  =sqrt  [ BE^2 + AB^2]  =sqrt [ 6^2 + 12^2] = sqrt 180  =  6sqrt (5)

Next  right triangle FCE   ..... [FC  = DC/3 = 4]

Area  1/2( CE)(FC)  = (1/2)(6)(4)   =12

And  FE  = sqrt [ CE^2 + FC^2]  =sqrt [ 6^2 + 4^2]  = sqrt 52  = 2sqrt (13)

Next triangle FDG...DF = 8  and  DG   = 3

Area= (1/2) (8)(3)  = 12

GF  =sqrt ( 8^2 + 3^2)  = sqrt 

Next triangle GAH  ....AH  = AE/ 3  = 2sqrt 5  =  sqrt (20)

We need to find GH  =  sqrt  [GA^2  - AH^2]  =  sqrt [ 9^2 - 20]  = sqrt (61)

Area of GAH  =(1/2) (AH)(GH)  = (1/2)(sqrt (20) (sqrt(61)  = sqrt (5)sqrt (61) = sqrt (305)

Area of FJG ...we know FJ  =(1/2)FE  = sqrt (13)

Area=  (1/2)(GF)(FJ)  =  (1/2)sqrt (73)sqrt (13)  = (1/2)sqrt (949)

Lastly  area of  JEH  ....JE  = sqrt (13)    HE =  (2/3)AE  = 4 sqrt (5)

Area  = (1/2)(JE)(HE) =  (1/2)sqrt (13) 4sqrt (5)  = 2 sqrt (65)

So....area  of  purple triangle  =

12^2  -  [ 36 + 12 + 12 + sqrt (305) + sqrt (949)/2 + 2sqrt (65) ]  ≈  35  units^2

If I didn't make any errors   !!!!   May 3, 2020