Mathematics Question and Solution
a² = 2²+2²-2*2*2cos120
a = √(8+½(8))
a = √(12)
a = 2√(3) units.
a is the side length of each of the 4 inscribed small regular hexagon.
sin30 = b/a
b = asin30
b = ½*2√(3)
b = √(3) units.
c = 4b+3a
c = 4(√(3))+3(2√(3))
c = 4√(3)+6√(3)
c = 10√(3) units.
c is the height of the big ascribed regular hexagon.
d = ½(c)
d = 5√(3) units.
sin60 = d/e
½√(3)e = 5√(3)
e = 10 units.
e is the side length of the big ascribed regular hexagon.
f² = 2(2√(3))²-2(2√(3))²cos120
f = 6 units.
Shaded Area is;
Area ascribed regular hexagon with side length 10 units - 4(area inscribed regular hexagon with side length 2√(3) units).
= (2(0.5*10*10sin120)+(10*10√(3))-4(2(0.5*2√(3)*2√(3)sin120)+(2√(3)*6))
= (50√(3)+100√(3))-4(6√(3)+12√(3))
= 150√(3)-4(18√(3))
= 150√(3)-72√(3)
= 78√(3) square units.
