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

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(A) Given that $\Delta ABC$ is an isosceles triangle with $AB = AC$. Let $AB = AC = x$ and $BC = y$.The perimeter is $AB + AC + BC = 50$,so $2x + y =

Vedclass · Accepted Answer

$120$

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(A) Given that $\Delta ABC$ is an isosceles triangle with $AB = AC$. Let $AB = AC = x$ and $BC = y$.The perimeter is $AB + AC + BC = 50$,so $2x + y = 50$,which implies $y = 50 - 2x$.Since $\overline{AM}$ is an altitude to the base $\overline{BC}$ in an isosceles triangle,it bisects the base. Thus,$BM = MC = y/2 = (50 - 2x)/2 = 25 - x$.In right-angled $\Delta ABM$,by the Pythagorean theorem: $AM^2 + BM^2 = AB^2$.Substituting the values: $15^2 + (25 - x)^2 = x^2$.$225 + 625 - 50x + x^2 = x^2$.$850 - 50x = 0$,so $50x = 850$,which gives $x = 17$.Now,find the base $y$: $y = 50 - 2(17) = 50 - 34 = 16$.The area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes BC 	imes AM$.Area $= \frac{1}{2} 	imes 16 	imes 15 = 8 	imes 15 = 120$.

In $\Delta ABC$,$\overline{AB} \cong \overline{AC}$ and $\overline{AM}$ is an altitude. If $AM = 15$ and the perimeter of $\Delta ABC$ is $50$,find the area of $\Delta ABC$.

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