Please help with this geometry problem! I'm totally stuck!
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.
+128406 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 [73]
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 !!!!
