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Answer :

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 \({m}^2\).

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