Calculating x.
a = 9+3
a = 12 units.
Therefore the ascribed plane shape (quadrilateral or irregular quadrilateral) is an equilateral triangle of side length 12 units with a subtraction of a s...
Notice.
The ascribed triangle is equilateral.
Let r be the radius of the inscribed half circle.
Calculating r.
sin60 = r/6
√(3)/2 = r/6
r = 3√(3) cm.
It implies, area orange (inscribed semi circl...
Let the side length of the regular octagon be 1 unit.
a = ⅛(180(8-2))
a= ⅛(180*6)
a = 135°
a is the single interior angle of the regular octagon.
2b² = 1²
b = √(1/2)
b = ½√(2) units.
c = 2b+1
c =...
Calculating red length.
Let r be the radius of the circle.
sin30 = a/2
a = 1 unit.
cos30 = b/2
b = √(3) units.
Or
b = ½√(2²+2²-2*2*2cos120)
b = ½√(8+4)
b = ½√(12)
b = ½*2√(3)
b = √(3) units.
c =...
8 = 0.5b²sin113.7
16 = b²sin113.7
b = √(16/sin113.7)
b = 4.1801537416 units.
Therefore;
x² = 2*4.1801537416²-2*4.1801537416²cos113.7
x = 6.9995991687 units.
x = 7 units.
Notice.
Since x+y = 1 cm.
It explains that the pentagon or irregular pentagon can be presented in form of a regular quadrilateral (square) with side length 1 unit.
Therefore;
Area red is;
1²
=...
a = 180-72-36
a = 180-108
a = 72°
Notice.
Triangle BCD is isosceles.
Let AD = BC = 1 unit.
Therefore, calculate b.
((b+1)/sin72) = (1/sin36)
bsin36+sin36 = sin72
bsin36 = sin72-sin36
b = (sin72-s...
Let a be the diameter of the smaller inscribed circle.
2a is the radius of the ascribed semi circle.
(2a)² = (0.5*3)²+b²
b² = 4a²-2.25
b = √(4a²-2.25) units.
(2a)² = (0.5*5)²+c²
c² = 4a²-6.25
c =...
a² = 2√(6)²
a² = 12
a = √(12)
a = 2√(3) units.
b = ½(a)
b = ½(2√(3))
b = √(3) units.
2c² = √(3)²
2c² = 3
c = √(3/2) units.
c =½√(6) units.
d = 2√(6)-½√(6)
d = ½(4√(6)-√(6))
d = ½(3√(6)) units.
d...
a = 90+20
a = 110°
sin20 = b/1
b = sin20 units.
b = 0.3420201433 units.
cos20 = c/1
c = cos20 units.
c = 0.9396926208 units.
Calculating d, the square side length observing similar plane shape si...
