Answer: B.pi/2
ABCD is a square
=> Δ ABD is a isosceles right triangle at A
using pythago theorem, we have:
AB² + AD² = BD²
=> 2AD² = 4
⇔ AD² = 2
⇒ AD = √2
through O, draw EF parallel to AD and BC
=> EF is diameter
because EF//AD => EF = AD = √2 (because circle O is inscribed the square ABCD)
=> the area of the circle is [tex]S=\frac{(\sqrt{2})^{2} }{4}.\pi =\frac{\pi }{2}[/tex]
Step-by-step explanation:
