question_answer

A field is in the shape of a trapezium whose parallel sides are 25m and 10m. The non parallel sides are 14m and 13m. Then the area of the field is

A)  190 \[{{m}^{2}}\]                  

B)  180 \[{{m}^{2}}\]         

C)  196 \[{{m}^{2}}\] 

D)  195 \[{{m}^{2}}\]

Correct Answer: C

Solution :

(c): From C, draw \[CE\parallel DA\]. Clearly, ADCE is a parallelogram having \[AD\parallel CE\] and \[DC\parallel AE\]such that \[~AD=13\] m and \[~DC=10\] m. \[\therefore \]\[AE=DC=10\]m and \[CE=AD=13\]m \[\Rightarrow \]\[BE=AB-AE=\left( 25-10 \right)m=15\]m Thus in \[\Delta BCE\], we have \[BC=14\]m, \[CE=13\]m and \[BE=15\]m Let?s be the semi - perimeter of \[\Delta BCE\]. Then, \[2s=BC+CE+BE=14+13+15=42\] \[\Rightarrow \]\[s=21\] Area of \[\Delta BCE=\sqrt{21\times (21-14)\times (21-13)\times (21-15)}\] \[\Rightarrow \]Area of \[\Delta BCE=\sqrt{21\times 7\times 8\times 6}\] \[\Rightarrow \]Area of \[\Delta BCE=\sqrt{{{7}^{2}}\times {{3}^{2}}\times {{4}^{2}}=84}\]m2 Also, Area of \[\Delta BCE=\frac{1}{2}\left( BE\times CL \right)\] \[\Rightarrow \]\[84=\frac{1}{2}\times 15\times CL\Rightarrow CL=\frac{168}{15}=\frac{56}{5}\] \[\Rightarrow \]Height of parallelogram\[ADCE=CL=\frac{56}{5}\]m \[\therefore \]Area of parallelogram ADCE = Base \[\times \]Height \[=AE\times CL=10\times \frac{56}{5}=112{{m}^{2}}\] Hence, Area of trapezium ABCE = Area of parallelogram ABCE = Area of parallelogram ADCE + Area of \[\Delta BCE\] \[=\left( 112+84 \right){{m}^{2}}=196\,{{m}^{2}}\]