question_answer

A square and an equilateral triangle have the same perimeter. If the diagonal of the square is \[12\sqrt{2}\] cm?  then the area of the triangle is

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 \[=\frac{16\times 16\times \sqrt{3}}{4}=64\sqrt{3}\,c{{m}^{2}}\]