Question : The centroid of an equilateral triangle $ABC$ is $G$ and $AB = 10 \:\operatorname{ cm}$. The length of $AG$ is:
Option 1: $3\frac{1}{3} \:\operatorname{ cm}$
Option 2: $\frac{10}{\sqrt3} \:\operatorname{ cm}$
Option 3: $10\sqrt3 \:\operatorname{ cm}$
Option 4: $\frac{1}{\sqrt3} \:\operatorname{ cm}$
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Correct Answer: $\frac{10}{\sqrt3} \:\operatorname{ cm}$
Solution : Given $AB =BC=AC= 10\ \text{cm}$ Let $AD$ be the median and $G$ be the centroid of an equilateral triangle $ABC$. $G$ divides AD in 2 : 1 ratio. $⇒AG=\frac{2}{2+1}×AD$ Also, D divides BC into equal parts. So, $BD=DC=\frac{10}{2}=5\ \text{cm}$ In $\triangle ABD$, from Pythagoras's theorem, we get, $AB^2=BD^2+AD^2$ $⇒10^2=5^2+AD^2$ $\therefore AD=\sqrt{75}=5\sqrt3$ Then, $AG=\frac{2}{2+1}×AD=\frac{2}{2+1}×5\sqrt3=\frac{10}{\sqrt3}\ \text{cm}$ Hence, the correct answer is $\frac{10}{\sqrt3}\ \text{cm}$.
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