In Delta ABC,overline{AB} cong overline{AC} a | vedclass

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(C) Given that $\Delta ABC$ is an isosceles triangle with $\overline{AB} \cong \overline{AC}$.In an isosceles triangle,the median to the base is also

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) Given that $\Delta ABC$ is an isosceles triangle with $\overline{AB} \cong \overline{AC}$.In an isosceles triangle,the median to the base is also the altitude to the base.Since $\overline{AD}$ is the median to the base $\overline{BC}$,$\overline{AD} \perp \overline{BC}$ and $D$ is the midpoint of $\overline{BC}$.Therefore,$BD = DC = \frac{BC}{2} = \frac{12}{2} = 6$.In the right-angled triangle $\Delta ABD$,by the Pythagorean theorem:$AB^2 = AD^2 + BD^2$$AB^2 = 8^2 + 6^2$$AB^2 = 64 + 36$$AB^2 = 100$$AB = \sqrt{100} = 10$.Thus,the length of $AB$ is $10$.

In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$ and $\overline{AD}$ is a median. If $BC = 12$ and $AD = 8$,find $AB$.

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