Question : In $\triangle$ABC, D is the midpoint of BC. Length AD is 27 cm. N is a point in AD such that the length of DN is 12 cm. The distance of N from the centroid of ABC is equal to:
Option 1: 3 cm
Option 2: 6 cm
Option 3: 9 cm
Option 4: 15 cm
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Correct Answer: 3 cm
Solution : AD = 27 cm ND = 12 cm Let O be the centroid. AO : OD = 2 : 1 ⇒ OD = $\frac{1}{3}$ AD = $\frac{1}{3}$ × 27 = 9 cm ⇒ ON = DN – OD = 12 – 9 = 3 cm Hence, the correct answer is 3 cm.
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