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 $
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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 $