UPSC CDS Math Quadrilateral and Polygon Questions

Let the distance between the parallel sides be h cm.ΔABE is an equilateral triangle.∠A = ∠B = ∠E = 60°Also, ∠EDC = ∠A=60° and ∠ECD = ∠B=60°  (Corresponding pair of angles)Therefore,ΔEDC is an equilateral triangleAE = BEDE = CE∴ AE- DE = BE - CE⇒AD = BC ...(1)Therefore, trapezium ABCD is an isosceles trapezium.⇒BD = AC  (Diagonals of isosceles trapezium are equal)Now ΔAPD ≅ ΔBQC (RHS congruence rule)∴ AP = BQ = b(say)Let CD = PQ = aArea of trapezium$= \frac{1}{2} \times AC \times OD + \frac{1}{2} \times AC \times OB$$= \frac{1}{2} \times AC \times BD$$= \frac{1}{2} BD^2$Using (1)$\therefore \frac{1}{2} BD^2 = 16$$\Rightarrow BD^2 = 32$Area of trapezium$\therefore (a+b) \times h = 16$$\Rightarrow a+b = \frac{16}{h}$In right ΔBPD,$h^2 + (a+b)^2 = BD^2$$\Rightarrow h^2 + \left(\frac{16}{h}\right)^2 = 32$Using option:$\Rightarrow h = 4\,\text{cm}$