In the figure ABCD is the field. Draw CF || DA and CG ⊥ AB.
DC = AF = 10 m, AD = FC = 13 m
For ∆BCF, a = 15 m, b = 14 m, c = 13 m
∴ s = a + b + c / 2 = 15 +14 +13 / 2 m = 21 m


Also, area of ∆BCF = 1/ 2 × base × height
= 1/ 2 × BF × CG
⇒ 84 = 1/ 2 × 15 × CG
⇒ CG = 84 x 2/ 15 m = 11.2 m
∴ Area of the trapezium = 1/ 2 × sum of the parallel sides × distance between them. = 1/ 2 × (25 + 10) × 11.2 m2
= 196 m2
Hence, area of the field = 196 m2 Ans.