Mathematics Question and Solution
Let a be the side of the inscribed square.
24² = a²+(2a)²
576 = a²+4a²
576 = 5a²
a = √(576/5)
a = 10.733126292 cm.
b = ½(24-10.733126292)
b = 6.633436854 cm.
b = AM = DR.
c = atan(6.633436854/(6+10.733126292))
c = 21.62463392714°
c = Angle BPQ.
tan21.62463392714 = d/10.733126292
d = 4.25488424946 cm.
d = MQ.
e = 0.5(10.733126292)+4.25488424946
e = 9.62144739546 cm.
e = MO.
f = 9.62144739546+12
f = 21.62144739546 cm.
f = DM.
g = atan(12/21.62144739546)
g = 29.03047057929°
g = Angle CMD.
Angle MCD = Angle MRN = 90-29.03047057929
= 60.96952942071°
h = 4.25488424946+10.733126292
h = 14.98801054146 cm.
h = MR.
sin60.96952942071 = i/14.98801054146
i = 13.1049432149 cm.
i = MN.
cos60.96952942071 = j/14.98801054146
j = 7.27330209225 cm.
j = NR.
k = 0.5(10.733126292)
k = 5.366563146 cm.
k = OR.
It implies;
Shaded Region Area is;
(0.5*7.27330209225*13.1049432149) - (0.5*5.366563146*7.27330209225sin60.96952942071)
= 30.59378383169 cm²
