Mathematics Question and Solution
Let the inscribed square side be 1 unit.
sin30 = a/1
a = ½ units.
a = 0.5 units.
a is the side length of one of the four congruent isosceles inscribed shaded triangle.
sin60 = b/1
b = ½√(3) units.
c = b-a
c = ½√(3)-½
c = ½(√(3)-1) units.
c = 0.3660254038 units.
d = a+a+c
d = 0.5+0.5+½(√(3)-1)
d = 1+½(√(3)-1)
d = ½(1+√(3)) units.
d = 1.3660254038 units.
d is the side length of the ascribed square.
Shaded Area is;
Area square with side 1 unit + 2(area square with side ½ units).
= 1²+2(0.5²)
= 1+½
= (3/2) square units.
Area Ascribed Square is;
d²
= (½(1+√(3)))²
= ¼(1+2√(3)+3)
= ¼(2√(3)+4)
= ½(2+√(3)) square units.
It implies;
Shaded Area Fraction is;
(3/2)÷½(2+√(3))
= 3/(2+√(3))
= 3(2-√(3))
= 0.8038475773
