A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Solution:

We know that, Heron's formula

Area of triangle = $\sqrt{s (s - a) (s - b) (s -c)}$

where, s is the semi-perimeter = half of the perimeter

and  a, b and c are the sides of the triangle 

DC = AF = 10 m, DA = CF = 13 m 

So, FB = 25 - 10 = 15 m

In ∆CFB, a = 15 m, b = 14 m, c = 13 m.

Semi Perimeter = s = (a $+$ b $+$ c)/2

= (15 $+$ 14 $+$ 13)/2

= 42/2

s = 21 m

By using above formula, 

Area of ∆CFB = $\sqrt{s (s - a) (s - b) (s -c)}$

= $\sqrt{21(21 - 15)(21 - 14)(21 - 13)}$

= $\sqrt{21 \times 6 \times 7 \times 8}$

= 84 m²

and 

Area of ∆CFB = 1/2 $\times$ base $\times$ height

84 = 1/2 $\times$ BF $\times$ CG

84 = 1/2$\times$ 15 $\times$ CG

CG = (84 $\times$2)/15

CG = 11.2 m 

Area of trapezium ABCD = 1/2 $\times$ sum of parallel sides$\times$ distance 

= 1/2 $\times$ (AB $+$ DC) $\times$ CG

= 1/2 $\times$ (25 $+$ 10) $\times$ 11.2

= 196 m²

Hence the area of the field is 196 m².