Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
tan30 = a/6
a = 2√(3) cm.
b = (6-2√(3)) cm.
sin30 = 2√(3)/c
c = 4√(3) cm.
6d+2√(3)d+ 4√(3)d = 6*2√(3)
(6+6√(3))d = 12√(3)
d = (72√(3)-216)/(-72)
d = (3-√(3)) cm.
d = 1.2679491924 cm.
d is the radius of the inscribed circle.
e = 6-d
e = (3+√(3)) cm.
Or
e = 4√(3)-(√(3)(3-√(3)))
e = 4√(3)-3√(3)+3
e = (3+√(3)) cm.
e = 4.7320508076 cm.
f² = 4.7320508076²+6²-2*6*4.7320508076cos60
f = 5.4772255751 cm.
(5.4772255751/sin60) = (6/sing)
g = 71.5650511768°
h = 90-g
h = 18.4349488232°
j = 120-71.5650511768
j = 48.4349488232°
k = 90-j
k = 41.5650511768°
l = 180-k-h
l = 120°
(5.4772255751/sin120) = (m/sin41.5650511768)
m = 4.1961524227 cm.
m = (3√(3)-1) cm.
(5.4772255751/sin120) = (n/sin18.4349488232)
n = 2 cm.
o = 6-n
o = 4 cm.
tan60 = p/(3-√(3))
p = (3√(3)-3) cm.
p = 2.1961524227 cm.
Therefore, purple area exactly in square cm is;
Area triangle with height and base 4 cm and (6-2√(3)) cm respectively + Area triangle with height and base 4.1961524227 cm and 2.1961524227 cm respectively.
= 0.5*4*(6-2√(3))+0.5*(3√(3)-1)*(3√(3)-3)
= 12-4√(3)+(0.5(27-12√(3)+3))
= 12-4√(3)+(15-6√(3))
= (27-10√(3)) cm²
= 9.6794919243 cm²
