question_answer

From a point in the interior of an equilateral triangle the perpendicular distances of the sides are \[\sqrt{3}\,cm,\] \[2\sqrt{3}\,cm\] and \[5\sqrt{3}\,cm.\] The perimeter (in\[c{{m}^{2}}\] ) of the triangle is

A) 64                    

B) 32

C) 48                                

D) 24

Correct Answer: A

Solution :

In AABC, O is the point from where there perpendicular line are drawn and also \[OP=\sqrt{3}\,cm,\] \[OQ=2\sqrt{3}\,cm\] and       \[OR=5\sqrt{3}\,cm.\] Let x be the side of the triangle. ar \[(\Delta AOB)=\frac{1}{2}\times OP\times AB\]             \[=\frac{1}{2}\times \sqrt{3}\times x\] ar \[(\Delta AOC)=\frac{1}{2}\times OQ\times AC\]             \[=\frac{1}{2}\times 2\sqrt{3}\times x\]             \[=\sqrt{3}x\] and ar \[(\Delta BOC)=\frac{1}{2}\times OR\times BC\]             \[=\frac{1}{2}\times 5\sqrt{3}\times x\] \[\therefore \] \[\frac{1}{2}\sqrt{3}x+\sqrt{3}x+\frac{5\sqrt{3}x}{2}=\frac{\sqrt{3}}{4}{{x}^{2}}\] \[\Rightarrow \] \[\frac{\sqrt{3}}{2}(1+2+5)=\frac{\sqrt{3}}{4}x\] \[\Rightarrow \] \[x=8\times 2=16\,cm\]  \[\therefore \] Perimeter \[=4\times 16=64\,cm\]