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.

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

Here, ABCD is a trapezium and CE || DA and \(CF\perp{AB}\).



Now, AB = 25 m, BC = 14 m, CD = 10 m, DA = 13 m, AE = 10 m and CE = 13m
\(\therefore \) EB = 25 - 10 = 15 m

Now, for \(\triangle{EBC}\), we have,
\(s = \frac{a + b + c}{2}\)
\(\therefore \) \(s = \frac{15 + 14 + 13}{2} = \frac{42}{2} = 21 m\)

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

= \(\sqrt{7 × 3 × 3 × 2 × 7 × 4 × 2}\)
= \(7 × 3 × 4 {m}^2\)
= \(84 {cm}^2\)

Also, Area of triangle \(\triangle{EBC}\)
= \(frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × 15 × CF
\(\therefore \) \(frac{1}{2}\) × 15 × CF = 84
\(\Rightarrow \) CF = \(\frac{84 × 2}{15} = \frac{168}{15} = 11.2m\)

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 \({m}^2\).