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# help

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Unit semicircle, with AB = 2.  Angle alpha = 30.  Find the area of the green region. Feb 11, 2020

#1
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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   Feb 11, 2020
#2
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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  Feb 12, 2020
edited by Dragan  Feb 12, 2020