In Delta ABC,mangle B = 90^{circ} and overlin | vedclass

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(C) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.Here,$\Delta ABC$ is a right-angled tri

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$20$

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(C) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.Here,$\Delta ABC$ is a right-angled triangle with $m\angle B = 90^{\circ}$.Since $\overline{BM}$ is the median to the hypotenuse $\overline{AC}$,we have $BM = \frac{1}{2} AC$.Given $BM = 12.5$,so $12.5 = \frac{1}{2} AC$,which implies $AC = 25$.Now,using the Pythagorean theorem in $\Delta ABC$:$AB^2 + BC^2 = AC^2$$15^2 + BC^2 = 25^2$$225 + BC^2 = 625$$BC^2 = 625 - 225 = 400$$BC = \sqrt{400} = 20$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is a median. If $AB = 15$ and $BM = 12.5$,find $BC$.

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