Mathematics Question and Solution
Let AB be 5 units.
Radius, r of the inscribed circle is;
r = 5*(2/5)
r = 2 units.
a = ⅛(180*6)
a = 135°
2b² = 25
b = ½(5√(2)) units.
c = 180-0.5(135)-45
c = 67.5°
sin22.5 = ½(5√(2))/d
d = 9.23879532511 units.
Where d is the side length of the regular triangle (purple area).
Area purple is;
0.5*9.23879532511*9.23879532511sin60
= 36.95994598698 square units.
Calculating Area Shaded.
e = 360-135-45-67.5-60
e = 52.5°
f² = 50-50cos135
f = 9.23879532511 units.
g = 52.5+0.5(45)
g = 75°
sin(22.5) = 2.5/h
h = 6.53281482438 units.
sin60 = 6.53281482438/i
i = 7.54344479484 units.
j = acos(6.53281482438/7.54344479484)
j = 30°
k = 90-j
k = 60°
Area Shaded is;
0.5*7.54344479484*9.23879532511sin75-0.5*25sin135-π*2*2
= 12.25361138127 square units.
Area Shaded ÷ Area Purple is;
12.25361138127÷36.95994598698
= 0.33153758898
≈ 0.33
