RRB Technician Grade I 19 Dec 2024 Shift 3 Paper

Concept:A quadrilateral with opposite sides equal and all interior angles $90^\circ$ is a rectangle.Explanation:Let the vertices be $A(9,0)$, $B(9,6)$, $C(-9,6)$, and $D(-9,0)$.Use the distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.Side $AB = \sqrt{(9-9)^2+(6-0)^2} = \sqrt{0+36} = 6$ units.Side $BC = \sqrt{(-9-9)^2+(6-6)^2} = \sqrt{(-18)^2+0} = 18$ units.Side $CD = \sqrt{(-9-(-9))^2+(0-6)^2} = \sqrt{0+36} = 6$ units.Side $DA = \sqrt{(9-(-9))^2+(0-0)^2} = \sqrt{18^2+0} = 18$ units.Thus, opposite sides are equal: $AB = CD = 6$ and $BC = DA = 18$.The horizontal and vertical sides are perpendicular, so each interior angle is $90^\circ$.Since opposite sides are equal and all angles are right angles, but adjacent sides are unequal, the figure is not a square; it is a rectangle.Answer:Rectangle, i.e., option B.