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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