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 $
BETA
