Mathematics Question and Solution
Calculating shaded area grey.
Let x be the side length of the ascribed regular hexagon.
½*a*asin120 = 12
√(3)a² = 48
a = √(48)/(√√(3))
a =4√(3)/(3)^(¼)
a = 4(3)^(½-¼)
a = 4(3)^(¼)
a = 5.26429605181 units.
a is the 2 equal lengths of the yellow area inscribed isosceles triangle.
It implies;
x² = 2a²-2a²cos120
x² = 2(5.26429605181)²-2(5.26429605181)²cos120
x = 9.11802822782 units.
Again, x is the side length of the ascribed regular hexagon.
sin30 = b/a
½ = b/5.26429605181
b = ½(5.26429605181)
b = 2.63214802591 units.
b is the height of the inscribed yellow triangle
c² = 2x²-2x²cos120
c² = 2(9.11802822782)²-2(9.11802822782)²cos120
c = 15.7928881554 units.
c is the height of the ascribed regular hexagon.
d = c-b
d = 15.7928881554-2.63214802591
d = 13.1607401295 units.
e² = d²+(½(x))²
e² = 13.1607401295²+(0.5(9.11802822782))²
e = 13.9280181809 units.
e is the two equal inscribed lengths.
f = ½(e)
f = 6.96400909044 units.
sing = d/e
g = asin(13.1607401295/13.9280181809)
g = 70.8933946489°
sin70.8933946489 = h/x
h = 9.11802822782sin(70.8933946489)
h = 8.61572683502 units.
cos70.8933946489 = j/x
j = 9.11802822782cos(70.8933946489)
j = 2.98457532452 units.
k = f-j
k = 6.96400909044-2.98457532452
k = 3.97943376592 units.
l = 90-g
l = 90-70.8933946489
l = 19.1066053511°
m = 120-90-l
m = 30-19.1066053511
m = 10.8933946489°
tan10.8933946489 = n/k
n = 3.97943376592tan(10.8933946489)
n = 0.76584238531 units.
o = h+n
o = 8.61572683502+0.76584238531
o = 9.38156922033 units.
Therefore, the indicated inscribed grey area is;
Area triangle with height 8.61572683502 units and base 2.98457532452 units + Area trapezoid with parallel sides 8.61572683502 units and 9.38156922033 units, and height 3.97943376592 units.
(0.5*8.61572683502*2.98457532452)+(0.5(8.61572683502+9.38156922033)*3.97943376592)
= 12.8571428573+35.809523809
= 48.6666666663 square units.
= ⅓(146) square units.
