Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let the side of the square be 1 unit.
Area square is;
1²
= 1 square unit.
1 = ½+¼-½(√(2))cosa
½(√(2))cosa = ¾-1
a = acos(-√(2)/4)
a = 110.70481105464°
b = 180-a
b = 69.29518894536°
c = 90-b
c = 20.70481105464°
(1/sin110.70481105464) = (1/2sind)
d = asin(sin110.70481105464/2)
d = 27.88556683609°
Area S is;
Area Semicircle with radius ½ unit - Area triangle with height ½ unit and base ½(sin(2*69.29518894536) units - 2(area sector with radius ½ unit and angle 20.70481105464° - Area sector with radius 1 unit and angle (2*27.88556683609°) + Area triangle with with height 1 unit and base sin(2*27.88556683609°) units.
= ½*¼(π) - ½*½*(½)sin(69.29518894536*2) - 2(20.70481105464*¼*π÷360) - (2*27.88556683609*π÷360) + 0.5sin(2*27.88556683609)
= 0.3926990817 - 0.08267972847 - 0.09034178098 - 0.48669495507 + 0.41339864235
= 0.14638125953 square units.
It implies;
Area S ÷ Area Square to 3 decimal places is;
0.14638125953 ÷ 1
= 0.14638125953
≈ 0.146
