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 \[=\,\frac{1}{2}\,\times (sum\,of\,parallel\,sides)\,kt\,height\] |
\[=\,\frac{1}{2}\,\times \,(25+10)\times \frac{56}{5}\] |
\[=\frac{1}{2}\times 35\times \frac{56}{5}\,=196{{m}^{2}}\] |
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