Question: $ \Delta ABC $ is an isosceles triangle and $ \overline{AB}=\overline{AC}=2a $ units, $ \overline{BC}=a $ unit. Draw $ AD\bot BC $ and find the length of $ \overline{AD}. $
Options:
A) $ \sqrt{15}aunits $
B) $ \frac{\sqrt{15}}{2},aunits $
C) $ \sqrt{17}aunits $
D) $ \sqrt{\frac{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 $