Question : G is the centroid of the equilateral triangle ABC. If AB = 10 cm, then the length of AG (in cm) is:
Option 1: $\frac{5 \sqrt3}{3}$
Option 2: $\frac{10 \sqrt3}{3}$
Option 3: $5 \sqrt3$
Option 4: $10\sqrt 3$
Correct Answer: $\frac{10 \sqrt3}{3}$
Solution :
AB = 10cm
Since AD is the perpendicular bisector of BC,
⇒ BD = 5cm and $\angle$ADB = 90°
⇒ AD = $\sqrt{\text{AB}^{2}-\text{BD}^{2}}$ = $\sqrt{10^{2}-5^{2}}$ = $\sqrt{75}$ = $5\sqrt{3}$ cm
Since G is the centroid, AG : GD = 2 : 1
⇒ AG = $\frac{2}{3}$AD = $\frac{2}{3}×5\sqrt{3}$ = $\frac{10}{\sqrt{3}}$ cm
Hence, the correct answer is $\frac{10\sqrt{3}}{3}$.
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