A) \[8\sqrt{3}\,\,cm\]
B) \[12\sqrt{3}\,\,cm\]
C) \[16\sqrt{3}\,\,cm\]
D) \[24\sqrt{3}\,\,cm\]
Correct Answer: D
Solution :
Since, PQR is an equilateral triangle Then, PL is also the median of . Similarly, RN and QM are also the median and O is the centroid. So, \[\frac{PO}{OL}=\frac{2}{1}\,\Rightarrow PL=\frac{PO}{2}=\frac{8}{2}=4\,cm\] Now, altitude of \[\Delta \,PQR=\frac{\sqrt{3}\,a}{2}\] (where, a = length of the side of equilateral triangle PQR) \[PO+OL=\frac{\sqrt{3}a}{2}\Rightarrow 8+4=\frac{\sqrt{3}a}{2}\] \[a=\frac{12\times 2}{\sqrt{3}}=\frac{24}{\sqrt{3}}\,\,cm\]You need to login to perform this action.
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