Let BF = CF be a.
It implies;
BC = 2a
BC is the side length of the regular pentagon.
Calculating a.
b = ⅕(180(5-2))-90
b = 108-90
b = 18°
b is angle BFG
Notice.
108° is the single interior angl...
Sir Mike Ambrose is the author of the question.
Area Blue Exactly in decimal cm² is;
Area triangle with height 3.84187454245 cm and base 5sin(180-atan(12/5)-atan(4/5))
= ½*3.84187454245*5sin(180-a...
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Calculating red length.
Observing Cosine Rule.
a² = 8²+12²-2*8*12cos40
a = 7.8050923711 units.
a is BD, the red length.
Observing Sine Rule.
(7.8050923711/sin40) = (12/sinb)
b = 81.2114041855°
b i...
Calculating AB, the side length of the ascribed square.
Let AB be a.
It implies;
a²+a² = 8²
2a² = 64
a² = 32
a = 4√(2) units.
Again a is AB, the side length of the ascribed square.
Therefore, r,...
Let the side of the inscribed square be a.
Calculating a.
tan60 = a/b
√(3) = a/b
b = a/√(3) units.
It implies;
b+b+a = 10
2(a/√(3))+a = 10
⅓(2√(3)a)+a = 10
2√(3)a+3a = 30
(2√(3)+3)a = 30
a = 30/...
Let the length of the ascribed regular hexagon be 2 units.
Therefore it area will be;
(2*2*6)/(4tan(180/6))
= 6/tan30
= 6√(3) square units.
The green areas is;
Area rectangle with length 2√(3) u...
Let the side of the ascribed regular hexagon be 2 units.
Area regular hexagon is;
(2*2*6)/(4tan(180/6))
= 6/tan30
= 6√(3) square units.
Area green is;
Area kite with perpendicular lengths 4 unit...
a = asin(1/6)
b = 2a
b = 2asin(1/6)
Where b is angle BAC
cos(2asin(1/6)) = c/6
c = ⅓(17) units.
Where c is AD.
d = 6-c
d = 6-⅓(17)
d = ⅓ units
Where d is CD.
sin(2asin(1/6)) = e/6
e = 1.97202659...
a = acos(3/5)
Where a is angle GCA.
tan(acos(3/5)) = b/3
b = 4 units.
Where b is AG.
c = atan(⅔)
Where c is angle GBE.
d = 180-acos(3/5)-atan(⅔)
d = 93.17983011986°
Where d is angle BEC.
(6/sin9...
