Answer :

Here, ABCD is a trapezium and AB || DC.

Here, ABCD is a trapezium and CE || DA and $CF?AB$.



Now, AB = 25 m, BC = 14 m, CD = 10 m, DA = 13 m, AE = 10 m and CE = 13m
$?$ EB = 25 - 10 = 15 m

Now, for $\mathrm{?}EBC$, we have,
$s=\frac{a+b+c}{2}$
$?$ $s=\frac{15+14+13}{2}=\frac{42}{2}=21m$

Now, Area of triangle
= $\sqrt{21(21?15)(21?14)(21?13)}$
(Since, Heron's formula [area = $\sqrt{s(s?a)(s?b)(s?c)}$])
= $\sqrt{21\times 6\times 7\times 8}$

= $\sqrt{7\times 3\times 3\times 2\times 7\times 4\times 2}$
= $7\times 3\times 4m^2$
= $84{cm}^2$

Also, Area of triangle $\mathrm{?}EBC$
= $frac12$ × Base × Height
= $\frac{1}{2}$ × 15 × CF
$?$ $frac12$ × 15 × CF = 84
$?$ CF = $\frac{84\times 2}{15}=\frac{168}{15}=11.2m$

Now, area of the trapezium ABCD
= $\frac{1}{2}$ × (Sum of parallel sides) × (Distance between parallel sides)
= $\frac{1}{2}$ × (AB + CD) × CF
= $\frac{1}{2}$ × (25 + 10) × 11.2
= $\frac{1}{2}$ × 35 × 11.2
= 35 × 5.65

Therefore, the area of the field is 196 $m^2$.