In an equilateral triangle ABC, G is the centroid and AB = 10 cm. What is the length of AG (in cm)?
In an equilateral triangle ABC, G is the centroid and AB = 10 cm. What is the length of AG (in cm)?
एक समबाहु त्रिभुज ABC का केन्द्रक G है तथा AB 10 सेमी है। AG की लम्बाई (सेमी में) है
Detailed Solution & Logic
\(\frac{10\sqrt{3}}{3}\)
Step 1: Find median of equilateral triangle
Side $a = 10$ cm
Median (height) of equilateral triangle:
$\frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3}$
Step 2: Centroid division
$AG = \frac{2}{3} \times \text{median}$So,
$AG = \frac{2}{3} \times 5\sqrt{3} = \frac{10\sqrt{3}}{3}$
Final Answer: $\frac{10\sqrt{3}}{3}$
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