In □ABCD; AB = DC and AD = BC also ∠A = 90° then □ABCD is a 

Correct option is (D) Square AB = \(\sqrt{(2-2)^2+(8-4)^2}\) = 4 units BC = \(\sqrt{(6-2)^2+(8-8)^2}\) = 4 units CD = \(\sqrt{(6-6)^2+(4-8)^2}\) = 4 units DA = \(\sqrt{(2-6)^2+(4-4)^2}\) = 4 units AC = \(\sqrt{(6-2)^2+(8-4)^2}\) = \(\sqrt{16+16}\) = \(4\sqrt2\) units BD = \(\sqrt{(6-2)^2+(4-8)^2}\) = \(\sqrt{16+16}\) = \(4\sqrt2\) units Since, AB = BC = CD...

Correct option is (B) 100°, 80°, 100° \(\because\) ABCD is a parallelogram. \(\therefore\) \(\angle B=\angle D\) = \(80^\circ\)   \((\because\) Opposite angles are equal in a parallelogram) Also \(\angle A+\angle B\) \(=180^\circ\)  (Sum of two consecutive angles is \(180^\circ\) in a parallelogram) \(\Rightarrow\) \(\angle...

Correct option is  A) 180 – x°

1.B) 80° 2. B) 45° 3. C) 90° 4. C) All the angles are equal in □ PQRS

Correct option is (B) 135° ABCD is a cyclic quadrilateral. \(\angle A\;\&\;\angle C\) are opposite angles in cyclic quadrilateral ABCD \(\therefore\) \(\angle A+\angle C\) \(=180^\circ\)    \((\because\) Sum of opposite angles in a cyclic quadrilateral is \(180^\circ)\) \(\Rightarrow\) \(\angle C\) \(=180^\circ-\angle A\) \(=180^\circ-45^\circ\) ...

Correct option is  B) 25 cm2 

Correct option is  B) 180°