A) \[64\sqrt{3}\,c{{m}^{2}}\]
B) \[32\sqrt{3}\,c{{m}^{2}}\]
C) \[24\sqrt{3}\,c{{m}^{2}}\]
D) \[24\sqrt{2}\,c{{m}^{2}}\]
Correct Answer: A
Solution :
Let x be the side of the square, then \[{{x}^{2}}+{{x}^{2}}=A{{C}^{2}}\] or \[2{{x}^{2}}={{(12\sqrt{2})}^{2}}\] or \[x=12\,cm\] Since the perimeters of square and triangle are equal, \[\therefore \] \[3y=4x\] or \[y=16\,cm\] \[\therefore \] Area of triangle \[=\frac{16\times 16\times \sqrt{3}}{4}=64\sqrt{3}\,c{{m}^{2}}\]You need to login to perform this action.
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