A) \[196\text{ }c{{m}^{2}}\]
B) \[148c{{m}^{2}}\]
C) \[200\text{ }c{{m}^{2}}\]
D) \[184\text{ }c{{m}^{2}}\]
Correct Answer: D
Solution :
: We have, \[AF=FE=ED=GD=GF=HG=AH\] Now, \[AF+FE=16\text{ }cm\] \[\Rightarrow \] \[AF+AF=16\text{ }cm\]\[\Rightarrow \] \[AF=8\text{ }cm\] \[\therefore \] Area of square AFGH \[=8\times 8=64\text{ }c{{m}^{2}}\] Now, area of rectangle HBCD \[=(15\times 16)c{{m}^{2}}\] \[=240\text{ }c{{m}^{2}}\] Area of shaded part of rectangle \[=240\div 2=120\text{ }c{{m}^{2}}\] \[\therefore \] Total shaded area \[=(64+120)\,c{{m}^{2}}\] \[=184\text{ }c{{m}^{2}}\]
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