question_answer

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.

A) \[~196\,c{{m}^{2}}\]                         

B)        \[~186\,c{{m}^{2}}\]             

C)        \[~169\,c{{m}^{2}}\]                         

D)        \[~199\,c{{m}^{2}}\]             

Correct Answer: A

Solution :

Let ABCD be the trapezium with sides AB = 25 cm, CD = 10 cm, AD = 14 cm, BC = 14 cm. We draw \[CE||AD\] \[\therefore \]Area of trapezium ABCD = area of parallelogram AECD + area of \[\Delta \Epsilon CB\] Now, In\[\Delta \Epsilon CB\] \[s=\frac{14+13+15}{2}=21\] \[\therefore \]Area of \[\Delta ECB\] \[=\sqrt{21(21-14)(21-13)(21-15)}\] \[=84\,c{{m}^{2}}\] Also, Area of \[\Delta EBC=\frac{1}{2}\times BE\times h\] \[\Rightarrow \]\[\frac{1}{2}\times 15\times h=84\] \[\therefore \]\[h=11.2\,cm\] \[\therefore \]Area of parallelogram AECD = \[AE\times h\] \[=10\times 11.2\] \[=112\,c{{m}^{2}}\] Hence, Area of trapezium ABCD \[=(112+84)c{{m}^{2}}=196\,c{{m}^{2}}\]