a² = (2r)²+r²
a² = 5r²
a = √(5)r units.
a is BD.
Where r is the radius of the circle.
b = ½(a)
b = ½(√(5)r) units because BD, points B and D are symmetry with respect to length AC.
tanc = 2r/r
c =...
Let the side length of the regular octagon be 1 unit.
a = ⅛(180(8-2))
a = ⅛(180*6)
a = 3*45
a = 135°
a is the single interior angle of the regular octagon.
b = ½(360-135-135)
b = ½(360-270)
b = ½(...
Sir Mike Ambrose is the author of the question.
Area ABC (yellow area) exactly is;
Area triangle with height ⅓√(265) cm and base ((5√(26)/6)sin84.3306169587) cm.
= ½*⅓√(265)*((5√(26)/6)sin84...
Calculating angle BAC, an interior angle of the inscribed triangle ABC.
0.5*15*18sinx = Area triangle ABC
135sina = 81
sinx = (81/135)
x = asin(81/135)
x = 36.8698976458°
x again, is angle BAC.
a²...
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/...
