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(D) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,the hypotenuse $AC$ can be calculated using the Pythagorean theorem:$AC^2 =

Vedclass · Accepted Answer

$12.5$

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(D) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,the hypotenuse $AC$ can be calculated using the Pythagorean theorem:$AC^2 = AB^2 + BC^2$$AC^2 = 15^2 + 20^2 = 225 + 400 = 625$$AC = \sqrt{625} = 25$.In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.Therefore,$BM = \frac{1}{2} 	imes AC = \frac{1}{2} 	imes 25 = 12.5$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is a median. If $AB = 15$ and $BC = 20$,find $BM$.

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