The adjoining figure shows a field with the measurement given in metres. Find the area of the field.

Calculating the area of triangle DXC,

$\text {Area of △DXC} = \dfrac{1}{2} \times CX \times DX \\[1em] = \dfrac{1}{2} \times 30 \times 12 \\[1em] = 15 \times 12 \\[1em] = 180 \text{ m}^2.$

Calculating the area of trapezium CXZB,

$\text {Area of trapezium CXZB} = \dfrac{1}{2} \times \text{(sum of parallel lines)} \times \text{distance between them} \\[1em] = \dfrac{1}{2} \times (CX + BZ) \times XZ \\[1em] = \dfrac{1}{2} \times (30 + 25) \times 15 \\[1em] = \dfrac{1}{2} \times 55 \times 15 \\[1em] = 27.5 \times 15 \\[1em] = 412.5 \text{ m}^2.$

Calculating the area of triangle AZB,

$\text {Area of △ AZB} = \dfrac{1}{2} \times \text{ base } \times \text{ height } \\[1em] = \dfrac{1}{2} \times BZ \times AZ \\[1em] = \dfrac{1}{2} \times 25 \times 10 \\[1em] = 25 \times 5 \\[1em] = 125 \text{ m}^2.$

From figure,

AD = 12 + 15 + 10 = 37 m.

Calculating the area of triangle AED,

$\text {Area of △AED} = \dfrac{1}{2} \times \text{ base } \times \text{ height } \\[1em] = \dfrac{1}{2} \times AD \times EY \\[1em] = \dfrac{1}{2} \times 37 \times 20 \\[1em] = 37 \times 10 \\[1em] = 370 \text{ m}^2.$

Total area = 180 + 412.5 + 125 + 370 = 1087.5 m².

Hence, area of the figure = 1087.5 m².