Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Let the side length of the regular ascribed pentagon be 1 unit.
a = ⅕*180(5-2)
a = 108°
a is the single interior angle of the regular pentagon.
Let the inscribed square side length be b.
(b/sin108) = (c/sin18)
c = 0.3249196962b units.
sin54 = (0.5b/d)
d = 0.6180339887b
It implies, b is;
c+d = 1
0.3249196962b+0.6180339887b = 1
0.9429536849b = 1
b = 1.060497473 units.
Again, b is the side length of the inscribed square.
d = 0.6180339887b
And b = 1.060497473 units.
d = 0.6180339887*1.060497473
d = 0.6554234832 units.
cos54 = e/d
cos54 = e/0.6554234832
e = 0.3852482574 units.
f = ½(b)
f = ½(1.060497473)
f = 0.5302487365 units.
let g be the side length of the inscribed regular hexagon.
h² = 2g²-2g²cos120
h = √(3)g units.
tan36 = h/j
j = √(3)g/(tan36)
j = 2.3839634169g units.
Therefore;
g+j = f
g+2.3839634169g = 0.5302487365
3.3839634169g = 0.5302487365
g = 0.156694583 units.
Again, g is the side length of the inscribed regular hexagon.
h = √(3)g
And g = 0.156694583 units.
h = √(3)*0.156694583
h = 0.271402979 units.
Therefore, area blue inscribed regular hexagon is;
Area rectangle with length 0.271402979 units and width 0.156694583 units + 2(area triangle with height 0.156694583 units and base 0.156694583sin120 units).
= (0.271402979*0.156694583)+2(½*0.156694583*0.156694583sin120)
= 0.0425273766+0.0212636883
= 0.0637910649 square units.
Calculating shaded area.
k = f-g
k = 0.5302487365-0.156694583
k = 0.3735541535 units.
It is;
Area triangle with height 0.271402979 units and base 0.3735541535 units - Area triangle with height 0.156694583 units and base 0.156694583sin120 units.
= ½(0.271402979*0.3735541535)-(0.5*0.156694583²sin120)
= 0.050691855-0.0106318442
= 0.0400600108 square units.
It implies;
Shaded Area ÷ Area Hexagon to 2 decimal places is;
0.0400600108÷0.0637910649
= 0.6279878053
≈ 0.63 to 2 decimal places.
