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 \[=4\times 16=64\,cm\]You need to login to perform this action.
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