Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Calculating exactly, hatched area ÷ orange area.
Let the ascribed square side length be 2 units.
Therefore, r, radius of the inscribed half circle is 1 unit.
a = (b+1) units.
b is the height of the inscribed orange area.
tanc = 2/1
c = atan(2)°
d = 180-2c
d = 180-2atan(2)
d = 2atan(½)°
Calculating b.
cosd = 1/(b+1)
cos(2atan(½)) = 1/(b+1)
⅗ = 1/(b+1)
3b+3 = 5
b = ⅔ units.
Again, b is the height of the inscribed orange area (trapezoid).
tand = e/1
tan(2atan(½)) = e/1
e = 4/3 units.
Or
e² = (1+(⅔))²-1²
e = 4/3 units.
Hatched Area is;
(½*(4/3)*1)-((2atan(½))π/360)
= ⅔-((atan(0.5))π/180)
= (120-atan(0.5)π)/180 square units.
= 0.20301905767 square units.
Calculating Orange Area.
f = 2-e
f = 2-(4/3)
f = ⅔ units.
cosd = f/g
cos(2atan(0.5)) = (⅔)/g
⅗ = 2/(3g)
9g = 10
g = 10/9 units.
Therefore, orange area is;
½((10/9)+2))*(⅔)
= ½*(28/9)*(⅔)
= 28/27 square units.
It implies, exactly, hatched area ÷ orange area is;
((120-atan(0.5)π)/180)÷(28/27)
= 27(120-atan(0.5)π)/(180*28)
= (360-3atan(0.5)π)/560
= 0.19576837703
