question_answer

The parallel sides of a trapezium shaped field are 25 m and 10 m and non parallel sides are 14 m and 13 m. What is the area of the field?

A) \[204{{m}^{2}}\]                    

B) \[196\,{{m}^{2}}\]

C) \[156\,{{m}^{2}}\]      

D)        \[144\,{{m}^{2}}\]

Correct Answer: B

Solution :

Sol. [b]

Let, AB and CD parallel side of the trapezium and BC and AD are unparallel sides.

AB = 25 m, CD = 10 m, AD = 14 m, BC = 13 m

Now DM and CN are perpendicular height of the trapezium ABCD.

Let, NB = a, then AM =15 - a

\[(\therefore \,AM\,+NB\,=\,25-10)\]

\[\therefore \,\,In\,\Delta ADM\]

\[{{h}^{2}}\,=\,{{14}^{2}}\,-\,{{(15\,-\,a)}^{2}}\]

\[{{h}^{2}}\,=\,196-(225+{{a}^{2}}-30a)\]

\[{{h}^{2}}\,=\,196-225-{{a}^{2}}+30a\]      (i)

\[In\,\Delta \,CNB\]

\[{{h}^{2}}\,{{13}^{2}}\,-\,{{a}^{2}}\]                                (ii)

From Eqs. (i) and (ii)

\[196-225-{{a}^{2}}+30a\,=169-{{a}^{2}}\]

\[30a\,=\,198\Rightarrow \,a=\,\frac{198}{30}\,=\frac{33}{5}\,m\]

Now using Eqs. (ii) \[{{h}^{2}}\,=169-{{\left( \frac{33}{5} \right)}^{2}}\]

\[=\,169-\frac{1089}{25}\,=\frac{4225-1089}{25}\]

\[h\,=\,\frac{3136}{25}\,=\,\frac{56}{5}\,m\]

\[\therefore \] Area of trapezium \[=\,\frac{1}{2}\,\times (sum\,of\,parallel\,sides)\,kt\,height\]

\[=\,\frac{1}{2}\,\times \,(25+10)\times \frac{56}{5}\]

\[=\frac{1}{2}\times 35\times \frac{56}{5}\,=196{{m}^{2}}\]