Mathematics Question and Solution
a² =2*12²
a = 12√(2) cm.
a = AC.
12b+12b+12√(2)b = 12*12
b = 144/(24+12√(2))
b = 12-6√(2) cm.
b = 3.51471862576 cm.
b = radius of the inscribed circle.
BF = ½(12√(2))
BF = 6√(2) cm.
c² = (6√(2))²+3.51471862576²-12√(2)*3.51471862576cos45
c = 6.49435320175 cm.
c = EF.
(6.49435320175/sin45) = (6√(2)/sind)
d = 112.5°
d = Angle BEF.
e = 2(180-112.5-45)
e = 45°
f = 112.5-90
f = 22.5°
cos22.5 = 12/g
g = 12.98870640351 cm.
g = EG.
FG = 12.98870640351-6.49435320175
FG = 6.49435320176 cm.
tan22.5 = h/12
h = 4.97056274848 cm.
i = 12-4.97056274848-3.51471862576
i = 3.51471862576 cm.
i = DG = radius of the inscribed circle.
tanj = 3.51471862576/12
j = 16.32494993689°
j = Angle DAG.
k² = 3.51471862576²+12²
k = 12.50412919872 cm.
k = AG.
l = 90-2(16.32494993689)
l = 57.35010012622°
m = 112.5-(90-16.32494993689)
m = 38.82494993689°
n = 180-38.82494993689-57.35010012622
n = 83.82494993689°
(12.50412919872/sin83.82494993689) = (o/sin57.35010012622)
o = 10.58970491347 cm.
p = 10.58970491347-6.49435320176
p = 4.09535171171 cm.
It implies;
Shaded Area Region is;
0.5*4.09535171171*6.49435320176sin45
= 9.40333949868 cm²
