Mathematics Question and Solution
Let EB be a.
Let AE be b.
It implies;
a/b = √(2) --- (1).
a+b = 1 --- (2).
From (1).
a = √(2)b --- (3).
Substituting (3) in (2) to get b.
√(2)b+b = 1
b(√(2)+1) = 1
b = (√(2)-1)/(2-1)
b = √(2)-1
b = 0.4142135624 units.
b is AE.
At (2).
a+b = 1
And b = 0.4142135624 units.
a = 1-0.4142135624
a = 0.5857864376 units.
a is EB.
tanc = 1/0.4142135624
c = 67.5°
c is angle AED.
d =½(c)
d = 33.75°
Let x be the small inscribed circle radius.
e = (0.4142135624-x) units.
Calculating x.
tan33.75 = x/(0.4142135624-x)
0.4142135624tan33.75-xtan33.75 = x
0.4142135624tan33.75 = x(1+tan33.75)
x = 0.1659106811 units.
Again, x is the small inscribed circle radius (r2).
Let y be the big inscribed circle radius.
Calculating y.
f = 180-67.5
f = 112.5°
f is angle BED.
g = ½(f)
g = 56.25°
h = (0.5857864376-y) units.
tan56.25 = y/(0.5857864376-y)
0.5857864376tan56.25-ytan56.25 = y
0.5857864376tan56.25 = y(1+tan56.25)
y = 0.3511533023 units.
Again, y is the small inscribed circle radius (r1).
Therefore;
r1 ÷ r2 is;
0.3511533023÷0.1659106811
= 2.1165201662
