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

(c): From C, draw CE∥DA. Clearly, ADCE is a parallelogram having AD∥CE and DC∥AEsuch that  AD=13 m and  DC=10 m. ∴AE=DC=10m and CE=AD=13m ⇒BE=AB−AE=(25−10)m=15m Thus in ΔBCE, we have BC=14m, CE=13m and BE=15m Let?s be the semi - perimeter of ΔBCE. Then, 2s=BC+CE+BE=14+13+15=42 ⇒s=21 Area of ΔBCE=21×(21−14)×(21−13)×(21−15)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√ ⇒Area of ΔBCE=21×7×8×6−−−−−−−−−−−√ ⇒Area of ΔBCE=72×32×42=84−−−−−−−−−−−−−−√m2 Also, Area of ΔBCE=12(BE×CL) ⇒84=12×15×CL⇒CL=16815=565 ⇒Height of parallelogramADCE=CL=565m ∴Area of parallelogram ADCE = Base ×Height =AE×CL=10×565=112m2 Hence, Area of trapezium ABCE = Area of parallelogram ABCE = Area of parallelogram ADCE + Area of ΔBCE =(112+84)m2=196m2