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# 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 $${m}^2$$.