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 m^{2}

= 196 m^{2}

Hence, area of the field = **196 m**^{2 }Ans.