question_answer

Lengths of the perpendiculars from a point in the interior of an equilateral triangle on its sides are 3 cm, 4 cm and 5 cm. Area of the triangle is

A)  \[48\sqrt{3}\,c{{m}^{2}}\]      

B)  \[54\sqrt{3}\,c{{m}^{2}}\]

C)  \[72\sqrt{3}\,c{{m}^{2}}\]                  

D)  \[80\sqrt{3}\,c{{m}^{2}}\]

Correct Answer: A

Solution :

Let the side of the equilateral triangle be x cm. \[\therefore \] \[\Delta AOB+\Delta BOC+\Delta COA=\Delta ABC\] \[\Rightarrow \]   \[\frac{1}{2}x\times 3+\frac{1}{2}\times x\times 4+\frac{1}{2}\times x\times 5\]             \[=\frac{\sqrt{3}}{4}{{x}^{2}}\] \[\Rightarrow \]   \[6=\frac{\sqrt{3}}{4}x\] \[\Rightarrow \] \[x=\frac{24}{\sqrt{3}}=8\sqrt{3}\] \[\therefore \] Area of \[\Delta ABC=\frac{\sqrt{3}}{4}\times sid{{e}^{2}}\]             \[=\frac{\sqrt{3}}{4}\times 8\sqrt{3}\times 8\sqrt{3}=48\sqrt{3}\] sq cm