Mathematics Question and Solution
a = 45-atan(3/4)
a = 45-36.86989764584
a = 8.13010235416°
Where a is angle DCF.
b² = 2(5²)
b = 5√(2) cm.
Where b is CF = AE.
c² = (5√(2))²+4²-10√(2)*4cos8.13010235416
c = 3.16227766017 cm.
Where c is DF.
(4/sind) = (3.16227766017/sin8.13010235416)
d = 10.30484646877°
Where d is angle DFC.
e = 45-10.30484646877
e = 34.69515353123°
Where e is angle AFD.
tan34.69515353123 = f/5
f = 3.46153846154 cm.
g = 90-34.69515353123
g = 55.30484646877°
h = atan(5/(5-3.46153846154))
h = 72.89727103096°
i = 90-72.89727103096
i = 17.10272896904°
tan(17.10272896904) = j/5
j = 1.53846153846 cm.
k = 180-atan(3/4)-72.89727103096
k = 70.2328313232°
(1.53846153846/sin70.2328313232) = (l/sin72.89727103096)
l = 1.5625 cm.
m = atan(3/4)
m = 36.86989764584°
Where m is angle DCA.
It implies;
Shaded Area Region is;
0.5*1.53846153846*4sin36.86989764584-0.5*1.5625*1.53846153846sin36.86989764584
= 1.125 cm²
= 1⅛ cm²
