Mathematics Question and Solution
a² = 12²-6²
a = 6√(3) cm.
a = 10.39230484541 cm.
a = circle's radius = AD = AG.
b² = 2*10.39230484541²-2*10.39230484541²cos120
b = 18 cm.
b = DG.
c = 12-10.39230484541
c = 1.60769515459 cm.
c = CD.
d² = 12²+1.60769515459²-24*1.60769515459cos60
d = 11.28239078631 cm.
d = BD.
(11.28239078631/sin60) = (1.60769515459/sine)
e = 7.088663081°
e = Angle CBD.
f = 60-7.088663081
f = 52.911336919°
f = Angle DBG.
(12/sing) = (10.39230484541/sin52.911336919)
g = 112.91133691899°
g = Angle BEG.
Angle BAE = 180-112.91133691899-52.911336919
= 14.17732616201°
Angle EAG = 180-14.17732616201
= 165.82267383799°
h² = 2*10.39230484541²-2*10.39230484541²cos165.82267383799
h = 20.6257399298 cm.
h = EG.
(20.6257399298/sin52.911336919) = ((12+10.39230484541)/sini)
i = 120°
i = Angle BEG.
j = 60-52.911336919
j = 7.088663081°
j = Angle BGE.
Therefore;
Angle DGE = 30-7.088663081
= 22.911336919°
k = 180-52.911336919
k = 127.088663081°
k = Angle DFG.
l = 180-127.088663081-22.911336919
l = 30°
l = Angle FDG.
(m/sin30) = (18/sin127.088663081)
m = 11.28239078631 cm.
m = FG
Therefore;
Shaded Area, Triangle DFG is;
0.5*11.28239078631*18sin22.911336919
= 39.53074360869 cm²
