A) \[300\text{ }c{{m}^{2}}\]
B) \[280\text{ }c{{m}^{2}}\]
C) \[270\text{ }c{{m}^{2}}\]
D) \[290\text{ }c{{m}^{2}}\]
Correct Answer: C
Solution :
We have, \[AD=20\text{ }cm\] \[\because \] P is mid point of AD. \[\therefore \] \[AP=PD=\frac{20}{2}=10\,cm\] Area of parallelogram \[ABCE=EC\times AP\] \[=18\times 10=180\,c{{m}^{2}}\] Area of \[\Delta \,EDC\] \[=\frac{1}{2}\times EC\times PD=\frac{1}{2}\times 18\times 10=90c{{m}^{2}}\] \[\therefore \] Required area \[=180+90=270\text{ }c{{m}^{2}}\]
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