Answer:
A₁ = x²* ( 1 - π /2) square units
Step-by-step explanation:
Let´s call the area of the region that is inside the square but not inside the circle A₁
And the area of the square A(s)
And the area of the circle A(c)
Then the area we are looking for is:
A₁ = A(s) - A(c)
If the circle has a radius r then
A(c) = π*r²
A(s) = x²
In the square diagonal of the square is d = √ x² + x² ; d = √2 * x
Half of this distance is d/2 = (√2 /2)*x
There is a right triangle with points:
o The center of the square (which is the center of the circle)
a Any of the corners of the square
b Middle point of a side of the square
According to that
r the radius of the circle is:
r² + (x/2)² = (d/2)²
r² + (x/2)² = [ (√2 /2)*x]²
r² = (2/4)*x² - x²/4
r² = (x²/2) - x² / 4
r² = (1/2)*x²
r = x/√2
Then the area of the circle is
A(c) = π * r²
A(c) = π * x²/2
And A₁ = x² - π * x²/2
A₁ = x²* ( 1 - π /2) square units or A₁ =[ ( 2 - π)/2 ]*x²
