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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]?

 Jan 4, 2021
 #1
avatar+114090 
0

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

 

cool cool cool

 Jan 5, 2021
 #2
avatar+824 
+4

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]?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

CF = 0.305            CE = 0.82          [CEF] = 1/8

 

[AEF] = 7/16  smiley

jugoslav  Jan 5, 2021
edited by jugoslav  Jan 6, 2021

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