Question

Can u explain how to solve this?

Answer 1

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 $S=\frac{(\sqrt{2})^{2} }{4}.\pi =\frac{\pi }{2}$

Step-by-step explanation: