A) \[59.86\,c{{m}^{2}}\]
B) \[50\,c{{m}^{2}}\]
C) \[59\,c{{m}^{2}}\]
D) \[53\,c{{m}^{2}}\]
Correct Answer: A
Solution :
Area of I & II part \[=2\times \frac{\sqrt{3}}{4}\times {{(2)}^{2}}=2\sqrt{3}\,c{{m}^{2}}=3.46\,c{{m}^{2}}\] Since, \[s=\frac{6+11+15}{2}=16\,cm\] [For III & IV part] \[\therefore \]Area of III & IV part \[=2\times \sqrt{16\times 10\times 15\times 1}\] \[=56.4\,c{{m}^{2}}\] Hence, total area of the mask \[=(3.46+56.4)\,c{{m}^{2}}=59.86\,c{{m}^{2}}\]
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