In Delta ABC,overline{AM} is an altitude. If | vedclass

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(D) Given that $\Delta ABC$ is an isosceles triangle with $AB = AC = 25$ and base $BC = 14$.Since $\overline{AM}$ is an altitude to the base $BC$,in a

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$24$

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(D) Given that $\Delta ABC$ is an isosceles triangle with $AB = AC = 25$ and base $BC = 14$.Since $\overline{AM}$ is an altitude to the base $BC$,in an isosceles triangle,the altitude from the vertex to the base also acts as the median.Therefore,$M$ is the midpoint of $BC$.Thus,$BM = MC = \frac{BC}{2} = \frac{14}{2} = 7$.In the right-angled triangle $\Delta ABM$,by the Pythagorean theorem:$AB^2 = AM^2 + BM^2$$25^2 = AM^2 + 7^2$$625 = AM^2 + 49$$AM^2 = 625 - 49 = 576$$AM = \sqrt{576} = 24$.Hence,the length of the altitude $AM$ is $24$.

In $\Delta ABC$,$\overline{AM}$ is an altitude. If $AB = AC = 25$ and $BC = 14$,find $AM$.

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