Draw CF ∥ AD and CG ⊥AB.
In quadrilateral ADCF,
CF ∥ AD [∵ By construction]
CD ∥ AF [∵ ABCD is a trapezium]
Therefore, ADCF is a parallelogram. So,
AD = CF = 13 m and CD = AF = 10 m [∵ Opposite sides of a parallelogram]
Therefore, BF = AB – AF = 25 – 10 = 15 m
Here, the Sides of triangle are a = 13 m, b = 14 m and c = 15 m.
So, the semi- perimeter of triangle
S = a + b + c/2 = 13 + 14 + 15/2 = 42/2 = 21 m.
Therefore, using Heron’s formula, area of triangle BCF = √s(s -a)(s -b)(s -c)
= √21(21 -13)(21 -14)(21 – 15)
= √21(8)(7)(6)
= √7056
= 84 m²
But, the area of triangle BCF = 1/2 BF × CG
So, 1/2 × BF × CG = 84
⇒ 1/2 × 15 × CG = 84
⇒ CG = (84×2/15) = 11.2 m
Therefore, area of trapezium ABCD = 1/2 × (AB + CD) × CG
= 1/2 × (25 + 10) × 11.2
= 35 × 5.6
196 m²
Hence, the area of the field is 196 m².