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

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(A) In $\Delta ABC$,$\overline{AD}$ is a median to the base $\overline{BC}$.Since $\Delta ABC$ is an isosceles triangle with $\overline{AB} \cong \ove

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) In $\Delta ABC$,$\overline{AD}$ is a median to the base $\overline{BC}$.Since $\Delta ABC$ is an isosceles triangle with $\overline{AB} \cong \overline{AC}$,the median $\overline{AD}$ to the base is also the altitude to the base.Therefore,$\overline{AD} \perp \overline{BC}$ and $\angle ADB = 90^{\circ}$.Let $AB = AC = x$. The perimeter of $\Delta ABC$ is $AB + AC + BC = 36$.$x + x + BC = 36 \implies BC = 36 - 2x$.Since $D$ is the midpoint of $\overline{BC}$,$BD = \frac{BC}{2} = \frac{36 - 2x}{2} = 18 - x$.In right-angled $\Delta ABD$,by the Pythagorean theorem:$AB^2 = AD^2 + BD^2$$x^2 = 12^2 + (18 - x)^2$$x^2 = 144 + 324 - 36x + x^2$$36x = 468$$x = 13$.Thus,$BD = 18 - 13 = 5$,so $BC = 2 	imes 5 = 10$.The area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes BC 	imes AD$.Area $= \frac{1}{2} 	imes 10 	imes 12 = 60$.

In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$ and $\overline{AD}$ is a median. If $AD = 12$ and the perimeter of $\Delta ABC$ is $36$,find the area of $\Delta ABC$.

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