Question : BL and CM are medians of $\triangle$ABC right-angled at A and BC = 5 cm. If BL=$\frac{3\sqrt5}{2}$ cm, then the length of CM is
Option 1: $2\sqrt5$ cm
Option 2: $5\sqrt2$ cm
Option 3: $10\sqrt2$ cm
Option 4: $4\sqrt5$ cm
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Correct Answer: $2\sqrt5$ cm
Solution : Given: BC = 5 cm. and BL=$\frac{3\sqrt5}{2}$ cm According to the question, $4(BL^2+CM^2)=5BC^2$ ⇒ $4[(\frac{3\sqrt5}{2})^2+CM^2]=5×5^2$ ⇒ $4×\frac{45}{4}+4CM^2=125$ ⇒ $4CM^2=125-45$ ⇒ $4CM^2=80$ ⇒ $CM^2=\frac{80}{4}$ ⇒ $CM=\sqrt20=2\sqrt5$ Hence, the correct answer is $2\sqrt5$ cm.
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