In Delta ABC,AB = AC = 37,BC = 70 and overlin | vedclass

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(B) In $\Delta ABC$,$AB = AC = 37$,which means $\Delta ABC$ is an isosceles triangle.Since $\overline{AM}$ is the median to the base $BC$,it is also t

Vedclass · Accepted Answer

$12$

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(B) In $\Delta ABC$,$AB = AC = 37$,which means $\Delta ABC$ is an isosceles triangle.Since $\overline{AM}$ is the median to the base $BC$,it is also the altitude to the base $BC$ in an isosceles triangle.Therefore,$\overline{AM} \perp \overline{BC}$ and $M$ is the midpoint of $BC$.Since $BC = 70$,$BM = MC = \frac{70}{2} = 35$.In the right-angled triangle $\Delta ABM$,by the Pythagorean theorem:$AB^2 = AM^2 + BM^2$$37^2 = AM^2 + 35^2$$1369 = AM^2 + 1225$$AM^2 = 1369 - 1225 = 144$$AM = \sqrt{144} = 12$.

In $\Delta ABC$,$AB = AC = 37$,$BC = 70$ and $\overline{AM}$ is a median. Find $AM$.

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