The following diagram shows a pentagonal field ABCDE in which the lengths of AF, FG, GH and HD are 50 m, 40 m, 15 m and 25 m respectively; and the lengths of perpendiculars BF, CH and EG are 50 m, 25 m and 60 m respectively. Determine the area of the field.

Area of figure ABCDE = Area of Δ ABF + Area of Δ AED + Area of Δ DHC + Area of trapezium BFHC

Area of Δ ABF = $\dfrac{1}{2}$ x b x h = $\dfrac{1}{2}$ x AF x BF

= $\dfrac{1}{2}$ x 50 x 50 m²

= 25 x 50 m²

= 1250 m²

Area of Δ AED = $\dfrac{1}{2}$ x b x h = $\dfrac{1}{2}$ x AD x GE

= $\dfrac{1}{2}$ x (AF + FG + GH + HD) x GE

= $\dfrac{1}{2}$ x (50 + 40 + 15 + 25) x 60 m²

= 130 x 30 m²

= 3900 m²

Area of Δ DHC = $\dfrac{1}{2}$ x b x h = $\dfrac{1}{2}$ x HD x CH

= $\dfrac{1}{2}$ x 25 x 25 m²

= 12.5 x 25 m²

= 312.5 m²

Area of trapezium BFHC = $\dfrac{1}{2}$ (sum of parallel sides) x distance between the parallel sides

= $\dfrac{1}{2}$ (BF + CH) x FH

= $\dfrac{1}{2}$ (BF + CH) x (FH + GH)

= $\dfrac{1}{2}$ (50 + 25) x (40 + 15) m²

= $\dfrac{1}{2}$ x 75 x 55 m²

= 37.5 x 55 m²

= 2062.5 m²

Thus, area of figure ABCDE = 1250 + 3900 + 312.5 + 2062.5 m²

= 7525 m²

Hence, the area is 7525 sq. m.