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.

Answers (1)

The trapezium field is shown below in figure:

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

field

Then in quadrilateral ADCF,

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

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

Therefore, ADCF is a parallelogram. 

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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