Quantitative Aptitude Ques 450

Question: In figure, ABC is a right triangle, right angled at B. AD and CE are the two medians drawn from A and C, respectively. If AC = 5 cm and $ AD=\frac{3\sqrt{5}}{2}cm. $ Find the length of CE.

Options:

A) $ 2\sqrt{5},cm $

B) 2.5 cm

C) 5 cm

D) $ 4\sqrt{2},cm $

Show Answer

Answer:

Correct Answer: C

Solution:

  • $ AC=5cm $
    $ \Rightarrow $ $ AD=\frac{3\sqrt{5}}{5}cm $ $ AE=BE $ and $ BD=CD $ $ AB^{2}=AC^{2}-BC^{2} $ $ =25-BC^{2} $ … (i)
    and $ AB^{2}=AD^{2}-BD^{2} $ $ ={{( \frac{3\sqrt{5}}{2} )}^{2}}-BD^{2} $ $ =\frac{45}{2}-\frac{BC^{2}}{4} $ From Eqs. (i) and (ii), we get $ BC^{2}=\frac{55}{3} $ Now, from Eqs. (i) $ AB^{2}=25-\frac{55}{3}=\frac{20}{3} $ Also, $ CE^{2}=BE^{2}+BC^{2}={{( \frac{1}{2}AB )}^{2}}+BC^{2} $ $ =\frac{1}{4}AB^{2}+BC^{2} $ $ [\because AB^{2}=25\cdot BC^{2}] $ $ =\frac{5}{3}+\frac{55}{3}=\frac{60}{3}=20 $

$ \therefore $ $ CE=2\sqrt{5},cm $

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