question_answer

\[\Delta ABC\] is an isosceles triangle and \[\overline{AB}=\overline{AC}=2a\,\,\text{units},\]\[\overline{BC}=a\,\,\text{units}.\]\[\overline{AD}\bot \overline{BC}\]and the length of \[\overline{AD}\] is

A) \[\sqrt{15}\,a\,\,\text{units}\]                  

B) \[\frac{\sqrt{15}}{2}a\,\,\text{units}\]

C) \[\sqrt{17}\,a\,\,\text{units}\]                  

D) \[\frac{\sqrt{17}}{2}a\,\,\text{units}\]

Correct Answer: B

Solution :

\[A{{D}^{2}}=A{{B}^{2}}-B{{D}^{2}}=4{{a}^{2}}-\frac{{{a}^{2}}}{4}\]

\[AD=\sqrt{\frac{15{{a}^{2}}}{4}}=\frac{a}{2}\sqrt{15}\,\text{units}\]