In a four-sides field, the length of the longer diagonal is 128 m.

The field, which is represented as ABCD, is given below.

The area of the field is the sum of the areas of triangles ABC and ADC.

Area of the triangle $\mathrm{ABC}=\frac{1}{2}(\mathrm{AC} \times \mathrm{BF})=\frac{1}{2}(128 \times 22.7)=1452.8 \mathrm{~m}^{2}$

Area of the triangle $\mathrm{ADC}=\frac{1}{2}(\mathrm{AC} \times \mathrm{DE})=\frac{1}{2}(128 \times 17.3)=1107.2 \mathrm{~m}^{2}$

Area of the field $=$ Sum of the areas of both the triangles $=1452.8+1107.2=2560 \mathrm{~m}^{2}$