Mathematics Question and Solution
b = √(a)
b = √(48)
b = √(16*3)
b = 4√(3) units.
b is the radius of the ascribed quarter circle.
c = ½(b)
c = 2√(3) units.
c is the radius of the inscribed half circle.
tan30 = d/c
⅓√(3) = d/(2√(3))
d = 2 units.
e = b-d
e = (4√(3)-2) units.
e = 4.92820323028 units.
e is the side length of the inscribed regular triangle.
f = ½(e)
f = (2√(3)-1) units.
tan60 = g/f
g = √(3)(2√(3)-1)
g = (6-√(3)) units.
g = 4.26794919243 units.
g shares the inscribed regular triangle to two equal right-angled triangles, g is the height (horizontally) of the inscribed regular triangle.
h = f+d
h = (2√(3)-1)+2
h = (2√(3)+1) units.
h = 4.46410161514 units.
h is the height of the inscribed yellow triangle.
j² = 4.92820323028²+2²-2*2*4.92820323028cos60
j = 4.29310850298 units.
(4.29310850298/sin60) = (2/sink)
k = 23.7939768872°
tan23.7939768872 = l/b
l = 4√(3)tan23.7939768872
l = 3.05483176319 units.
m = g-l
m = (6-√(3))-3.05483176319
m = 1.21311742924 units.
m is the base of the inscribed yellow triangle.
Therefore the required yellow inscribed right-angled triangle is;
½(hm)
= 0.5*4.46410161514*1.21311742924
= 2.70773973761 square units.
