MAT Exam 2022 Paper

Concept:In an isosceles trapezium, the non-parallel sides are equal, and the perpendiculars from the shorter base to the longer base cut the longer base symmetrically.Explanation:Let $AB \parallel CD$, with $AD = BC$ as the equal non-parallel sides.Given $\angle A = 45^\circ$ and one non-parallel side $AD = 10\sqrt{2}$.Area of $\triangle ABD$ is $200$ sq. units.Using the area formula:$$200 = \frac{1}{2} \times AD \times AB \times \sin 45^\circ.$$Substitute $AD = 10\sqrt{2}$:$$200 = \frac{1}{2} \times 10\sqrt{2} \times AB \times \frac{\sqrt{2}}{2}.$$$$200 = 5AB.$$$$AB = 40.$$Draw $DP \perp AB$ and $CQ \perp AB$.In right $\triangle ADP$, $\angle A = 45^\circ$, so $AP = DP$.Also:$$AP^2 + DP^2 = AD^2.$$$$2AP^2 = (10\sqrt{2})^2 = 200.$$$$AP = 10.$$By symmetry of the isosceles trapezium, $BQ = AP = 10$.Therefore, the other parallel side is:$$CD = AB - AP - BQ = 40 - 10 - 10 = 20.$$Sum of the parallel sides:$$AB + CD = 40 + 20 = 60.$$Answer:$60$, i.e., option A.