Quantitative Aptitude Ques 293

Question: $ \Delta ABC $ is an isosceles triangle and $ \overline{AB}=\overline{AC}=2aunits, $ $ \overline{BC}=aunits. $ $ \overline{AD}\bot \overline{BC} $ and the length of $ \overline{AD} $ is

Options:

A) $ \sqrt{15},aunits $

B) $ \frac{\sqrt{15}}{2}aunits $

C) $ \sqrt{17},aunits $

D) $ \frac{\sqrt{17}}{2}aunits $

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ AD^{2}=AB^{2}-BD^{2}=4a^{2}-\frac{a^{2}}{4} $ $ AD=\sqrt{\frac{15a^{2}}{4}}=\frac{a}{2}\sqrt{15},units $