Mathematics Question and Solution
b² = 12²+12²
b² = 144+144
b² =288
b = 12√(2) cm.
Where b is AE = AC.
c² = 12²+6²
c = 6√(5) cm.
Where c is AM.
d = b-c
d = (12√(2)-6√(5))
d = 3.55415488348 cm.
Where d is EM.
e = 45-atan(1/2)
e = 18.43494882292°
Where e is angle MAC.
f = 45+18.43494882292
f = 63.43494882292°
Where f is angle EAB.
sin63.43494882292 = g/(12√(2))
g = 15.17893276881 cm.
cos63.43494882292 = h/(12√(2))
h = 7.5894663844 cm.
i = 12-7.5894663844
i = 4.4105336156 cm.
j = atan(15.17893276881/4.4105336156)
j = 73.79775141036°
Where j is angle ABE.
k = 90-73.79775141036
k = 16.20224858964°
Where k is angle CBE.
l = 180-45-16.20224858964
l = 118.79775141036°
Where l is angle BNC.
(12/sin118.79775141036) = (m/sin16.20224858964)
m = 3.82089210645 cm.
Where m is CN.
2n² = 3.82089210645²
n = 2.70177871865 cm.
Where n is BR.
tan16.20224858964 = 2.70177871865/o
o = 9.29822128133 cm.
Where o is NR.
p = 12-9.29822128133
p = 2.70177871867 cm.
q = 6-2.70177871865
q = 3.29822128135 cm.
r² = 12²+3.29822128135²
r = 12.44500958701 cm.
Where r is MR.
s = atan(12÷3.29822128135)
s = 74.63164469797°
Where s is angle ARM.
t = 90-74.63164469797
t = 15.36835530203°
Where t is angle MRN.
u = 180-45-15.36835530203
u = 119.63164469797°
(9.29822128133/sin119.63164469797) = (v/sin45)
v = 7.56404622865 cm.
It implies;
Area Shaded Region is;
Area triangle with height 9.29822128133 cm and base 12.44500958701sin15.36835530203 cm - Area triangle with height 9.29822128133 cm and base 7.56404622865sin15.36835530203 cm.
= (0.5*9.29822128133*12.44500958701sin15.36835530203) - (0.5*9.29822128133*7.56404622865sin15.36835530203)
= 6.0139523566 cm²
