A) \[~40\,c{{m}^{2}}\]
B) \[~20\,c{{m}^{2}}\]
C) \[~80\text{ }c{{m}^{2}}\]
D) \[~60\text{ }c{{m}^{2}}\]
Correct Answer: C
Solution :
Joint HF, Then \[HF||AB||CD\] \[\Rightarrow \]\[ar(\Delta FGH)=\frac{1}{2}(HFCD)\] ?.(i) \[\Rightarrow \]\[ar(\Delta \Epsilon FH)=\frac{1}{2}(ABFH)\] ?.(ii) From (i) and (ii),we get \[ar(EFGH)=\frac{1}{2}(ABCD)\] \[\Rightarrow \]\[40=\frac{1}{2}(ABCD)\] \[\Rightarrow \]\[80\,c{{m}^{2}}=ABCD\]You need to login to perform this action.
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