\( ABCD \) 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 \( a + b \)
+128631 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
