\[\Delta ABC\]is an isosceles triangle and \[\overline{AB}=\overline{AC}=2a\]units, \[\overline{BC}=a\]unit. Draw \[AD\bot BC\]and find the length of \[\overline{AD}.\] |
[SSC (CGL) 2014] |
A) \[\sqrt{15}a\,\,\text{units}\]
B) \[\frac{\sqrt{15}}{2}\,a\,\,\text{units}\]
C) \[\sqrt{17}a\,\,\text{units}\]
D) \[\sqrt{\frac{17}{2}\,}a\,\,\text{units}\]
Correct Answer: B
Solution :
\[A{{D}^{2}}=A{{B}^{2}}-B{{D}^{2}}=4{{a}^{2}}-\frac{{{a}^{2}}}{4}\] |
\[AD=\sqrt{\frac{15{{a}^{2}}}{4}}=\frac{a}{2}\sqrt{15}\,\,\text{units}\] |
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