A) \[21\text{ }c{{m}^{2}}\]
B) \[~18\,c{{m}^{2}}\]
C) \[~16\text{ }c{{m}^{2}}\]
D) \[~14\text{ }c{{m}^{2}}\]
Correct Answer: D
Solution :
Form the given figure D,E,F are mid-points sides BC, AC, AB. Then, \[ar(\Delta \Alpha \Beta C)=ar[\Delta \Alpha F\Epsilon +\Delta FED+\Delta EDC+FBD]\] Also, \[ar\,\Delta \Alpha FE=ar\Delta FED=ar\Delta EDC=ar\Delta FBD\] \[\Rightarrow \]\[ar(\Delta \Alpha F\Epsilon )=\frac{ar(\Delta \Alpha \Beta C)}{4}=\frac{28}{4}=7\,c{{m}^{2}}\] Now, Area of parallelogram AEDF \[=ar(\Delta \Alpha \Epsilon F+\Delta EFD)=(7+7)c{{m}^{2}}=14c{{m}^{2}}\]
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