Mathematics Question and Solution
Calculating Area Blue.
a = 6+4
a = 10 units.
a is OQ.
b = a+4
b = 10+4
b = 14 units.
tanc = 6/10
c = atan(3/5)°
c is angle OQP.
tand = 6/14
d = atan(3/7)°
e = 180-c
e = 180-atan(3/5)
e = (90+atan(5/3))°
e = 149.036243468°
f = 180-e-d
f = 180-149.036243468-atan(3/7)
f = 7.76516601835°
g = 180-2d
g = 180-2atan(3/7)
g = 133.602818973°
h = e-g
h = 149.036243468-133.602818973
h = 15.433424495°
tanj = 14/6
j = atan(7/3)°
k = 180-2j
k = 180-2atan(7/3)
k = 46.3971810273°
tanl = 10/6
l = atan(5/3)°
m = 180-2l
m = 180-2atan(5/3)
m = 61.9275130641°
n = m-k
n = 61.9275130641-46.3971810273
n = 15.5303320368°
o² = 2(6)²-2(6)²cos46.3971810273
o = 4.72703158295 units.
p² = 2(6)²-2(6)²cos61.9275130641
p = 6.17394906513 units.
It implies;
Area Blue is;
Area triangle with height 6 units and base 4 units - Area triangle with height 4 units and base 4sin133.602818973 units - Area sector with radius 4 units and angle 15.433424495° - Area sector with radius 6 units and angle 15.5303320368° + Area triangle with height 6 units and base 6sin15.5303320368 units - Area triangle with height 4.72703158295 units and base 6.17394906513sin7.76516601835 units.
(0.5*6*4)-(0.5*4*4sin133.602818973)-(15.433424495π*4*4÷360)-(15.5303320368π*6*6÷360)+(0.5*6*6sin15.5303320368)-(0.5*4.72703158295*6.17394906513sin7.76516601835)
= 2.02085645248 square units.
