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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