Question : ABC is an isosceles triangle and $\overline{AB} = \overline{AC} = 2a$ units and $\overline{BC} = a$ unit. Draw $\overline{AD} \perp \overline{BC}$, then find the length of $\overline{AD}$.
Option 1: $\sqrt {15} a$
Option 2: $\frac{\sqrt {15}}{2} a$
Option 3: $\sqrt {17} a$
Option 4: $\frac{\sqrt {17}}{2} a$
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Correct Answer: $\frac{\sqrt {15}}{2} a$
Solution :
By Pythagoras theorem,
AD = $\sqrt{(2a)^2 - (\frac{a}{2})^2}$ = $\sqrt{\frac{16a^2-a^2}{4}}$ = $\frac{\sqrt {15}}{2} a$
Hence, the correct answer is $\frac{\sqrt {15}}{2} a$.
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