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

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(B) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.Here,$\overline{BE}$ is the median to t

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.Here,$\overline{BE}$ is the median to the hypotenuse $\overline{AC}$ because $\angle B = 90^{\circ}$.Therefore,$BE = \frac{1}{2} AC$.Given $BE = 8.5$,we have $8.5 = \frac{1}{2} AC$,which implies $AC = 17$.In $\Delta ABC$,by the Pythagorean theorem,$AB^2 + BC^2 = AC^2$.Substituting the known values: $15^2 + BC^2 = 17^2$.$225 + BC^2 = 289$.$BC^2 = 289 - 225 = 64$.$BC = \sqrt{64} = 8$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BE}$ is a median. If $AB = 15$ and $BE = 8.5$,find $BC$.

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