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

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(A) $1$. In $\Delta ABC$,since $m\angle B = 90^{\circ}$,we can find the length of $BC$ using the Pythagorean theorem: $AB^2 + BC^2 = AC^2$.$2$. Substi

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$2.4$

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(A) $1$. In $\Delta ABC$,since $m\angle B = 90^{\circ}$,we can find the length of $BC$ using the Pythagorean theorem: $AB^2 + BC^2 = AC^2$.$2$. Substituting the given values: $3^2 + BC^2 = 5^2$,which gives $9 + BC^2 = 25$.$3$. Therefore,$BC^2 = 25 - 9 = 16$,so $BC = 4$.$4$. The area of $\Delta ABC$ can be calculated in two ways: $	ext{Area} = \frac{1}{2} 	imes AB 	imes BC$ or $	ext{Area} = \frac{1}{2} 	imes AC 	imes BM$.$5$. Equating the two expressions: $\frac{1}{2} 	imes 3 	imes 4 = \frac{1}{2} 	imes 5 	imes BM$.$6$. Simplifying: $12 = 5 	imes BM$,which leads to $BM = \frac{12}{5} = 2.4$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AB = 3$ and $AC = 5$,find the length of $BM$.

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