Mathematics Question and Solution
Calculating green inscribed triangle area.
a² = 2(3√(2))²
a² = 2*18
a = √(36)
a = 6 units.
a is AB.
b = ½(360-90)
b = ½(270)
b = 135°
b is angle ABC.
c² = 6²+(4√(2))²-2*6*4√(2)cos135
c = 2√(29) units.
c = 10.7703296143 units.
c is AC.
2d² = (2√(29))²
d² = 2*29
d = √(58) units.
d = 7.6157731059 units.
d is AO = CO the radius of the ascribed quarter circle.
sine = (0.5*6)/7.6157731059
e = 23.1985905135°
f = 2e
f = 46.3971810271°
g = 90-f
g = 43.6028189729°
g is angle BOC.
h = ½(180-g)
h = 68.1985905135°
h is angle OBC = angle OCB.
tanj = (3/4)
j = atan(3/4)°
j is angle BCH.
k = h-j
k = 31.3286928677°
k is OCH.
l = √((3√(2))²+(4√(2))²)
l = 5√(2) units.
l = 7.0710678119 units.
l is CH.
Therefore, green triangle inscribed area is;
½*CH*CO*sin(angle OCH)
= 0.5*5√(2)*√(58)sin31.3286928677
= 14 square units.
