# Q9.    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.

The trapezium field is shown below in figure:

Drawing line CF parallel to AD and a line perpendicular to AB, we obtain

$CF || AD$                         ............................ $\left [ \because\ by\ construction \right ]$

$CD || AF$                        ............................  $\left [ \because ABCD\ is\ a\ trapezium \right ]$

So, $AD = CF = 13\ m$  and  $CD = AF = 10\ m$

$\left ( \because Opposite\ sides\ of\ a\ parallelogram \right )$

Therefore, $BF = AB-AF = 25-10 = 15\ m$

Now, the sides of the triangle;

$a = 13\ m,\ b =14\ m\ and\ c = 15\ m.$

So, the semi-perimeter of the triangle will be:

$s= \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21\ m$

Therefore, the area of the triangle can be found by using Heron's Formula:

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

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

$= \sqrt{21(8)(7)(6)}$

$= \sqrt{7056} = 84\ m^2.$

Also, the area of the triangle is given by,

$Area = \frac{1}{2}\times BF\times CG$

$\Rightarrow \frac{1}{2}\times BF\times CG = 84\ m^2$

$\Rightarrow \frac{1}{2}\times 15\times CG = 84\ m^2$

Or,

$\Rightarrow CG = \frac{84\times2}{15} = 11.2\ m$

Therefore, the area of the trapezium ABCD is:

$= \frac{1}{2} \times (AB+CD)\times CG$

$= \frac{1}{2} \times (25+10)\times 11.2$

$= 35\times5.6$

$= 196\ m^2$

Hence, the area of the trapezium field is $196\ m^2.$

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