Let the side length of the regular pentagon be 1 unit.
z = √(2-2cos108)
z = 1.61803398875 units.
sin18 = y/1
y = 0.30901699437 units.
cos18 = x/1
x = 0.9510565163 units.
Area Regular Pentagon is;...
sina = 7/25
a = asin(7/25)°
tan(asin(7/25)) = b/25
b = 7.29166666667 cm.
c = 25-7.29166666667
c = 17.70833333333 cm.
cos(asin(7/25)) = 25/d
d = 25.96150997149 cm.
(7/25) = e/17.70833333333
e = 4...
Let the side length of the inscribed yellow square be 1 unit.
Area yellow square is;
1²
= 1 square unit.
Side length of the small regular pentagon is 1 unit.
a² = 2-2cos108
a = 1.61803398875 unit...
Sir Mike Ambrose is the author of the question.
a² = 2²+2²
a = 2√(2) units.
2b² = 4
b = √(2) units.
c = 3-√(2) units.
Let the side of the inscribed regular pentagon be d.
Calculating d.
e = ⅕(1...
EG = 3AG
Let AG = 1 unit.
Therefore;
EG = 3 units.
The side length of the regular pentagon is;
AG+EG = 1+3 = 4 units.
AE = 4 units.
It implies, Area Blue is;
0.5*2*4sin108
= 4sin108
= 3.80422606...
Sir Mike Ambrose is the author of the question.
Let the side of the blue square (length CD) be 1 unit.
Area Blue is;
1²
= 1 square unit.
a = 180-20-40
a = 120°
(b/sin40) = (1/sin20)
b = 1.87938524...
Let the side length of the regular pentagon be 2 units.
(CF)² = 5-4cos108
CF = 2.49721204096 units.
CF is the side length of the blue regular triangle and radius of the sector CGF.
Area Blue is;...
Let AO be 1 unit.
AB will be 2 units.
Therefore, radius r of the ascribed semi circle is;
= 3 units.
Area semi circle is;
½(3²)π
= ½(9π) square units.
tana = (1/3)
a = atan(⅓)°
b = (180-2atan(½...
Sir Mike Ambrose is the author of the question.
Let the side of the inscribed square be a.
Calculating a.
¾ = a/b
b = ⅓(4a)
It implies;
½*a*((⅓(4a))+3a) = 78
⅙(13a²) = 78
⅙(a²) = 6
a = 6 cm.
b =...
Calculating x.
Let the two congruent lengths of the plane shape be 1 unit.
(a/sin11) = (1/sin22)
a = 0.50935834748 units.
(b/sin147) = (1/sin22)
b = 1.45389601942 units.
c = 1+a
c = 1.5093583474...
