Mathematics Question and Solution
Notice!
The side length of the small inscribed square is twice the side length of the big inscribed square.
Let the side length of the small inscribed square be a units
Therefore, the side length of the big inscribed square is 2a units.
Radius of the ascribed semi circle is 1 unit.
It implies;
1² = (2a)²+a²
1 = 5a²
a = √(1/5)
a = ⅕√(5) units.
a is the side length of the small square.
Let b be the radius of the inscribed shaded circle.
c = (1-b) units.
d = (⅕√(5)+b)
Therefore, calculating b.
2(⅕√(5)+b)² = (1-b)²
2(⅕+⅕(2√(5)b)+b²) = 1-2b+b²
⅖+⅘√(5)b+2b² = 1-2b+b²
b²+(⅘√(5)b+2b)+⅖-1 = 0
b²+⅕(4√(5)+10)b-⅗ = 0
(b+⅕(2√(5)+5))² = ⅗+1.894427191²
(b+1.894427191)² = 4.188854382
b = -1.894427191±√(4.188854382)
It implies;
b = -1.894427191+√(4.188854382)
b = -1.894427191+2.0466690944
b = 0.1522419034 units.
Again, b is the radius of the inscribed shaded circle.
Area shaded inscribed circle is;
π(0.1522419034)²
= 0.0728145689 square units.
