Mathematics Question and Solution
Let the radius of the quarter circle be 1 unit.
2a² = 1
a² = ½
a = ½√(2) units.
a is the side length of the inscribed square.
b = 1-a
b = 1-½√(2)
b = ½(2-√(2)) units.
b is the side length of the green square G.
Area green square is;
b²
= (½(2-√(2)))²
= ½(3-2√(2)) square units.
Calculating Area R, Red Area.
(½√(2)+c)²+c² = 1²
½+√(2)c+2c² = 1
4c²+2√(2)c-1 = 0
(c+¼√(2))² = ¼
(c+¼√(2))² = ¼+(¼√(2))²
c = -¼√(2)±√(6/16)
c = -¼√(2)±¼√(6)
Therefore;
c ≠ -¼√(2)-¼√(6)
c = -¼√(2)+¼√(6)
c = ¼(√(6)-√(2)) units.
Again, c is the side length of the reg inscribed square.
Area G, red inscribed square is;
c²
= (¼(√(6)-√(2)))²
= ¼(2-√(3)) square units.
Therefore;
Area G ÷ Area R is;
½(3-2√(2)) ÷ ¼(2-√(3))
= (6-4√(2))/(2-√(3))
= 1.2806373753
