Mathematics Question and Solution
Let the side of the ascribed regular hexagon be 1 unit.
a = ⅙(180(6-2))
a = 120°
a is the single interior angle of the ascribed regular hexagon.
b² = 1²+1²-2*1*1cos120
b = √(3) units.
b is the height of the ascribed regular hexagon.
c = ½(b)
c = ½(√(3)) units.
cos60 = d/1
d = ½ units.
e = 2d+1
e = 2*½+1
e = 2 units.
It implies, area ascribed regular hexagon is;
2(½(1+2)*½(√(3)))
= 2(½*3*½*√(3))
= ½(3√(3)) square units.
= 2.5980762114 square units.
Calculating area inscribed red regular hexagon.
Let the radius of the inscribed bigger quarter circle be f.
f² = 2g²
g² = f²/2
g = ½(√(2)f) units.
g is the radius of the inscribed smaller quarter circle.
g² = 2h²
(½(√(2)f))² = 2h²
½(f²) = 2h²
h² =¼(f²)
h = ½(f) units.
h is half the radius, f of the bigger inscribed quarter circle.
It implies;
f+½(f) = b
½(3f) = √(3)
3f = 2√(3)
f = ⅓(2√(3)) units.
Again, f is the radius of the inscribed bigger quarter circle.
Let j be the side length of the red inscribed regular hexagon.
cos60 = (0.5j/k)
½ = (0.5j/k)
k = j units.
It implies;
f² = j²+(2j)²-2*2j²cos120
(⅓(2√(3)))² = 5j²+2j²
⅓(4) = 7j²
j² = 4/21
j = √(4/21)
j = 2/√(21)
j = (2√(21)/21) units.
j = 0.4364357805 units.
Again, j is the side length of the red inscribed regular hexagon.
I² = 0.4364357805²+0.4364357805²-2*0.4364357805*0.4364357805cos120
l = 0.7559289461 units.
Therefore, area red inscribed regular hexagon is;
(0.7559289461*0.4364357805)+2(0.5*0.4364357805²sin120)
= 0.3299144396+2(0.0824786099)
= 0.3299144396+0.1649572198
= 0.4948716594 square units.
Therefore, shaded fraction is;
Area inscribed red regular pentagon ÷ Area ascribed regular hexagon
= 0.4948716594÷2.5980762114
= 0.1904761905
≈ 0.2 to 1 decimal place.
