Construction: →Draw BE ∥ AD such that D−E−C →Draw BM ⊥ DC such that D−E−M−C ▱ABED is parallelogram, →AD=BE=13 m →AB=DE=10 m →BC=14 m →DC=DE+EC (∵D−E−C) ∴EC=25−10=15 m →In ΔBEC →2S=13+14+15 →S=21 m →Area of ΔBEC →A=√s(s−a)(s−b)(s−c) =√21(21−13)(21−14)(21−15)=√21×8×7×6=84 m² →Area of ΔBCE=1/2×BM×EC →BM=84×2/15=11.2 cm →Area of ▱ABED=11.2×10=112 m² →Area of the field ▱ABCD=84+112 m²=196 m²