Sir Mike Ambrose is the author of the question.
Let the side of the regular pentagon be 1 unit.
Area regular pentagon is;
½(5)*(1/(2tan(36)))
= 1.72047740059 square units.
a = ⅕(180*3)
a = 108°
(1...
Let the side length of the regular hexagon be 1 unit.
a² = 2-2cos120
a = √(3) units.
Notice;
AB = AC
Calculating y.
It implies;
((√(3)-1)/sin75) = (y/sin45)
y = (4-2√(3)) units.
x = 1-y
x = 1...
Sir Mike Ambrose is the author of the question.
Notice;
BC = 2 units.
a = ⅛(180*6)
a = 135°
b = 2+2(2sin45)
b = 4.82842712475 units.
c = b+2
c = 6.82842712475 units.
Let AB be d.
Calculating d....
Let the regular pentagon side be 1 unit.
Let the width and length of the inscribed red rectangle be a and b respectively.
(1/sin72) = (c/sin54)
c = 0.85065080835 units.
d = 180-54-108
d = 18°
(d...
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Sir Mike Ambrose is the author of the question.
tan30 = 1/a
a = √(3) units.
b = a-1
b = (√(3)-1) unit.
It implies;
Area Shaded in square units exactly as a single fraction is;
Area triangle with...
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...
