Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Showing that brown area ÷ smaller octagon area = a/b.
Let the side length of the smaller regular octagon be 1 units.
a = ⅛*180(8-2)
a = 135°
a is the single interior angle of the regular octagons.
2b² = 1²
b² = ½
b = ½√(2) units.
c = 1+2b
c = 1+2*½√(2)
c = (1+√(2)) units.
c is the height of the smaller regular octagon.
Area smaller regular octagon is;
(2(½(1+(1+√(2))))*½√(2))+(1*(1+√(2)))
= (2+√(2))*½√(2)+(1+√(2)
= ½(2√(2)+2)+(1+√(2))
= 2(1+√(2)) square units.
Calculating Area Brown.
2d² = x²
d² = x²/2
d = ½√(2)x units.
x is the side length of the bigger regular octagon.
e = x+d
e = x+½√(2)x
e = ½(2+√(2))x units.
Notice.
c = e
Therefore, calculating x.
(1+√(2)) = ½(2+√(2))x
2+2√(2) = (2+√(2))x
x = (2+2√(2))/(2+√(2))
x = √(2) units.
Again, x is the side length of the regular bigger octagon.
Recall.
d = ½√(2)x
And x = √(2) units.
d = ½√(2)*√(2)
d = 1 unit.
f = x+2d
f = √(2)+2(1)
f = (2+√(2)) units.
f is the height of the bigger regular octagon.
g = f-1
g = (2+√(2))-1
g = (1+√(2)) units.
h = ½(2+√(2)) units.
tanj = (2+√(2))/√(2)
j = atan((2+√(2))/√(2))°
tan(atan((2+√(2))/√(2))) = (½(2+√(2)))/k
2k(2+√(2)) = 2√(2)+2
(2+√(2))k = (1+√(2))
k = ½√(2) units.
Brown Area is;
(½(2+√(2))*√(2))-(½*½(2+√(2))*½√(2))+(½*√(2)*(1+√(2))sin45)
= (1+√(2))-(⅛(2√(2)+2))+(½(1+√(2))
= 1+√(2)-¼√(2)-¼+½+½√(2)
= ¼(4+4√(2)-√(2)-1+2+2√(2))
= ¼(5+5√(2))
= ¼(5(1+√(2)) square units.
It implies;
Showing brown area ÷ smaller octagon area = a/b, where a and b are integers is;
¼(5(1+√(2)) ÷ (2(1+√(2)))
= 5/(4*2)
= ⅝
a = 5 units.
b = 8 units.
