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\( ABCD \) is a rectangle with sides \( AB = 2 \) and \( BC = 1 \). \( EF \) is a circular arc with point \( B \) as its centre. Also, point \( E \) is a midpoint of the segment \( CD \). If the green region can be represented as \( \frac{1}{a}(b - \pi) \), where \( a \) and \( b \) are both positive integers, find the value of \( a + b \)

 

 Jan 27, 2021
 #1
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Let A  = (0,0)

B  =  (2,0)

E  =(1,1)

 

BE    =  the radius of the  circle  =   sqrt  [ (2-1)^2  +(1-0)^2  ] = sqrt (2)

 

F =  (2 -sqrt (2)  ,  0)

 

EF  =  sqrt   [ ( 2 - sqrt (2)  - 1)^2  + 1^2  ]  =   sqrt  [ ( 1  -sqrt (2))^2  + 1  ) ]  = 

 

sqrt [ 1 - 2sqrt (2) + 2  + 1]  =  sqrt  (4  -2sqrt (2) )  =  sqrt [ 4  -sqrt (8) ] 

 

Using the  Law of  Cosines

 

EF^2   = BE^2 + BF^2  - 2 ( BE * BF)  cos EBF

 

4 -sqrt (8)  =  2  + 2  -  2(2) cos EBF

 

-sqrt (8)  =   -4 cos EBF

 

sqrt (8)  / 4 = cos EBF

 

2sqrt ( 2)   /4  =cos  EBF

 

sqrt (2)   / 2 = cos EBF

 

So EBF  = 45°

 

Area of  the  circle  =  pi  (sqrt (2))^2  ( 1/8)   =  pi/4

 

Area of shaded  region  =  area of  rectangle  -area of  circle  =

 

AB * BC  -  pi/4  =

 

2 * 1    -  pi/4   =

 

2  - pi/ 4   =

 

(1/4) ( 8  - pi)

 

a =  4     b  =   8

 

And their  sum  =  12

 

 

 

cool cool cool

 Jan 27, 2021

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