Let ABCD be a square with side length 1. Points E and F lie on line BC and line CD, respectively, in such a way that angle EAF=45 degrees If [CEF]=1/8, what is the value of [AEF]?
+128475 See the following :
Angle EAB =22.5°
Angle AEB = 67.5°
Using the Law of Sines
BE /sin22.5 = 1/sin AEB
BE = sin 22.5 / sin 67.5
BE = sin22.5/ cos22.5
BE = tan 22.5 = (1 -cos 45)/ sin 45 = (1 - 1/sqrt (2)) sqrt (2) = sqrt (2) - 1
So the area of triangle AEB = (1/2) ( AB) ( BE) = (1/2) (1) (sqrt (2) - 1)
And this = area of triangle ADF
So [ AEF ] = [ area of square ] - [ CEF] - 2 [ AEB] =
1 - (1/8) - ( (sqrt (2) - 1 ) =
2 - 1/8 - sqrt (2) =
15/8 - sqrt (2) ≈ .461 units^2
+1639 Let ABCD be a square with a side length of 1. Points E and F lie on line BC and line CD, respectively, in such a way that angle EAF=45 degrees. If [CEF] = 1/8, what is the value of [AEF]?
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CF = 0.305 CE = 0.82 [CEF] = 1/8
[AEF] = 7/16 
