Unit semicircle, with AB = 2. Angle alpha = 30. Find the area of the green region.
Connect AD
Then triangle ADB is right with AB = 2 , AD =1 and BD = √3
And in triangle AEB, angle AEB = 120°
And using the law of sines
EB / sin 30 = AB /sin 120
EB /(1/2) = 2 / (√3/2)
2EB = 4/√3
EB = 2/√3 = (2/3)√3
So DB - EB = DE = √3 - (2/3)√3 = √3/3
And by symmetry DE = CE
And angle DEC =120°
So the area of triangle DEC = (1/2) DE^2 * sin (120°) = (1/2) (√3/3)^2 * (√3/2) = √3/12 (1)
Next....let F be the midpoint of AB
Connect DF and CF
Angle DFC = 60°
DF = CF = 1
So the area of sector DFC = (1/2) (DF) (CF) * pi/3 = (1/2) (1)^2 * (pi/3) = pi/6 (2)
And the area of triangle DFC = (1/2)(1) sin (60°) = √3/4 (3)
So....the green area = (1) + [ (2) - (3) ] =
√3/12 + [ pi/6 - √3/4 ] =
pi/6 + √3/12 - 3√3/12 =
2pi /12 + √3 /12 - 3√3/12 =
[ 2pi - 2√3 ] pi - √3
__________ = _________ ≈ .235 units^2
12 6
Unit semicircle, with AB = 2. Angle alpha = 30. Find the area of the green region.
There are 2 sectors: ADG and BCG, and 2 right angled triangles: DFG, and CFE.
Area of both sectors is: [(AB/2)² * pi] /3 = 1.04719755 u²
Area of triangle DFG =(DF * GF) /2 = 0.21650635 u²
Area of triangle CFE = (CF * EF) /2 = 0.072168783 u²
Area of semicircle is: pi / 2 = 1.570796327 u²
Area of the green (yellow) region is: A = 0.234923643 u²