Mathematics Question and Solution
Let the radius of the ascribed half circle be 1 unit.
a² = 2x²
a = √(2)x units.
b²+1² = a²
b = √(2x²-1) units.
c = 1-b
c = (1-√(2x²-1)) units.
d = ½(c)
d = ½(1-√(2x²-1)) units.
e²+x² = 2²
e = √(4-x²) units.
It implies;
x ~ 2
½(1-√(2x²-1)) ~ x
Cross Multiply.
(1-√(2x²-1)) = x²
1-x² = √(2x²-1)
(1-x²)² = 2x²-1
1-2x²+x⁴ = 2x²-1
x⁴-4x²+2 = 0
Let x² = p.
p²-4p = -2
(p-2)² = -2+(2)²
(p-2)² = 2
p = 2±√(2)
It implies;
p = (2+√(2)) units.
Or
p = (2-√(2)) units.
Recall.
x² = p
For p = (2+√(2))
x ≠ √(2+√(2)) units.
x ≠ 1.84775906502 units.
For p = (2-√(2))
x = √(2-√(2)) units.
x = 0.76536686473 units.
x is the side length of the inscribed square.
Calculating r.
Recall.
d = ½(1-√(2x²-1))
And x = 0.76536686473 units.
d = 0.5(1-√(2(0.76536686473)²-1))
d = 0.29289321881 units.
f = (1-r) units.
(1-r)² = r²+g²
g² = 1-2r+r²-r²
g = √(1-2r) units.
h = (r-0.29289321881) units.
j = (√(1-2r)-0.29289321881) units.
Therefore;
r² = (r-0.29289321881)²+(√(1-2r)-0.29289321881)²
r² = r²-0.58578643762r+0.08578643762+1-2r-0.58578643762√(1-2r)+0.08578643762
2.58578643762r = 1.17157287525-0.58578643762√(1-2r)
0.58578643762√(1-2r) = 1.17157287525-2.58578643762r
(0.58578643762)²(1-2r) = (1.17157287525-2.58578643762r)²
0.3431457505-0.686291501r = 1.37258300202-6.05887450301r+6.68629150098r²
6.68629150098r²-5.37258300201r+1.02943725152 = 0
It implies;
r = 0.31545 units.
r is the radius of the inscribed circle.
Therefore;
x(inscribed square side length) ÷ r(inscribed circle radius) is;
x÷r
= 0.76536686473÷0.31545
= 2.42626997854
≈ 2.43 to 2 decimal places.
