Mathematics Question and Solution
Calculating the required inscribed orange triangle area.
a = ½(3)
a = 1.5 units.
a is the radius of the inscribed circle.
tanb = 3/4
b = atan(3/4)°
tan(atan(3/4)) = c/1.5
3/4 = c/1.5
4c = 4.5
c = 1.125 units.
d = (180-atan(4/3))°
e = a-c
e = 1.5-1.125
e = 0.375 units.
(1.5/sin(180-atan(4/3))) = (0.375/sinf)
f = 11.5369590328°
g = 180-2f
g = 180-2(11.5369590328)
g = 156.926081934°
h² = 2(1.5)²-2(1.5)²cos156.926081934
h = 2.93938769134 units.
h is the longest length of the inscribed orange triangle.
j = (180-atan(4/3))+f
j = (180-atan(4/3))+11.5369590328
j = 138.406856679°
k² = 2(1.5)²-2(1.5)²cos2(1.5)²-2(1.5)²cos138.406856679
k = 2.80454076851 units.
k is the longer length of the inscribed orange triangle.
l = ½(180-j)
l = 0.5(180-138.406856679)
l = 20.7965716605°
m = f+l
m = 11.5369590328+20.7965716605
m = 32.3335306933°
m is the interior angle opposite the long length of the inscribed orange triangle.
Therefore, the required inscribed orange triangle area is;
0.5*h*k*sinm
= 0.5*2.93938769134*2.80454076851sin32.3335306933
= 4.12181630741sin32.3335306933
= 2.2045407685 square units.
