Find the area of the field shown in Fig. 20.39 by dividing it into a square,

The given figure can be divided into a square, a parallelogram and a trapezium as shown in following figure:

From the above figure:

Area of the figure $=$ (Area of square AGFM with sides $4 \mathrm{~cm}$ ) $+$ (Area of rectangle ME DN with length $8 \mathrm{~cm}$ and width $4 \mathrm{~cm}$ )+(Area of trapezium NDCB with parallel sid es $8 \mathrm{~cm}$ and $3 \mathrm{~cm}$ and perpendicular height $4 \mathrm{~cm})=(4 \times 4)+(8 \times 4)+[12 \times(8+3) \times(4)]=$ $16+32+22=70 \mathrm{~cm} 2$