Question : In $\triangle$ ABC, $\angle$ C = 90$^{\circ}$. M and N are the midpoints of sides AB and AC, respectively. CM and BN intersect each other at D and $\angle$ BDC = 90$^{\circ}$. If BC = 8 cm, then the length of BN is:
Option 1: $6 \sqrt{3} {~cm}$
Option 2: $6 \sqrt{6} {~cm}$
Option 3: $4 \sqrt{6} {~cm}$
Option 4: $8 \sqrt{3} {~cm}$
Correct Answer: $4 \sqrt{6} {~cm}$
Solution : Given: BN and CM are medians of $\triangle$ ABC, D is the centroid of the triangle. D will divide BN in the ratio 2 : 1 $\therefore$ BD : DN = 2 : 1 Let BD = 2$x$ and DN = $x$ ⇒ BN = 3$x$ In right-angled $\triangle$ CNB, CD perpendicular to BN ⇒ DC 2 = BD × BN ⇒ DC 2 = $2x \times x$ ⇒ DC = $x\sqrt2$ Now, in $\triangle$ CDB BC 2 = CD 2 + BD 2 ⇒ 8 2 = 2$x^2$ + $(2x)^2$ ⇒ 64 =6$x^2$ ⇒ $x = \frac{8}{\sqrt6}$ ⇒ $BN = 3x = 3 \times \frac{8}{\sqrt6} = 4\sqrt 6$ Hence, the correct answer is $4\sqrt 6~cm$.
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