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

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(D) In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$,which means $\Delta ABC$ is an isosceles triangle.In an isosceles triangle,the median to the b

Vedclass · Accepted Answer

$108$

Vedclass · Answer

(D) In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$,which means $\Delta ABC$ is an isosceles triangle.In an isosceles triangle,the median to the base is also the altitude to the base.Therefore,$\overline{AD} \perp \overline{BC}$,so $\overline{AD}$ is the height of the triangle with respect to base $\overline{BC}$.Area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$Area $= \frac{1}{2} 	imes BC 	imes AD$Area $= \frac{1}{2} 	imes 18 	imes 12$Area $= 9 	imes 12 = 108$.

In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$ and $\overline{AD}$ is a median. If $AD = 12$ and $BC = 18$,then the area of $\Delta ABC$ is.............

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