Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Calculating Area Green.
Area green exactly in cm² is;
Area ABCD - Area trapezoid with parallel sides (9-2√(3)) cm and (24√(3)-39) cm, and height 6 cm.
= (54-6√(3)) - ½*6((9-2√(3))+(24√(3)-39))
= (54-6√(3)) - 3(22√(3)-30)
= 54-6√(3)-66√(3)+90
= 144-72√(3)
= 72(2-√(3)) cm²
Or
Calculating Area Green.
Let a be BC.
tan30 = 12/b
b = 12√(3) cm.
b is the height of the ascribed right-angled triangle.
tan30 = a/c
c = √(3)a cm.
d = 6+c
d = (6+√(3)a) cm.
tan30 = e/(6+√(3)a)
e = (6+√(3)a)/√(3)
e = (2√(3)+a) cm.
e is AD.
Calculating a, length BC.
It implies;
½(a+(2√(3)+a))*6 = 54-6√(3)
6a+6√(3) = 54-6√(3)
6a = 54-12√(3)
a = (9-2√(3)) cm.
a = 5.53589838486 cm.
Again, a is BC.
Recall.
c = √(3)a cm.
And a = (9-2√(3)) cm.
c = √(3)(9-2√(3))
c = (9√(3)-6) cm.
f = c+6
f = (9√(3)-6)+6
f = 9√(3) cm.
g = b-f
g = 12√(3)-9√(3)
g = 3√(3) cm.
h = 6+g
h = (6+3√(3)) cm.
tanj = (9-2√(3))/(6+3√(3))
j = atan((9-2√(3))/(6+3√(3)))
tanj = k/(3√(3))
(9-2√(3))/(6+3√(3))= k/(3√(3))
(6+3√(3))k = (27√(3)-18)
k = (27√(3)-18)/(6+√(3))
k = (24√(3)-39) cm.
k = 2.56921938165 cm.
Recall.
e = (2√(3)+a) cm.
e is AD.
And a = (9-2√(3)) cm.
e = 2√(3)+a
e = 2√(3)+9-2√(3)
e = AD = 9 cm.
l = e-k
l = 9-(24√(3)-39)
l = (48-24√(3)) cm.
l = 6.43078061835 cm.
l is the base of the required green triangle area.
Area Green is;
½*6l
= ½*6(48-24√(3))
= 72(2-√(3)) cm²
= 19.292341855 cm²
