In Delta ABC,AB = 6 , cm,BC = 8 , cm,and AC = | vedclass

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(B) First,observe that $AB^2 + BC^2 = 6^2 + 8^2 = 36 + 64 = 100 = 10^2 = AC^2$. Since the sum of the squares of two sides equals the square of the thi

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$4.8$

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(B) First,observe that $AB^2 + BC^2 = 6^2 + 8^2 = 36 + 64 = 100 = 10^2 = AC^2$. Since the sum of the squares of two sides equals the square of the third side,$\Delta ABC$ is a right-angled triangle with the right angle at $B$. The area of $\Delta ABC$ can be calculated as $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes BC = \frac{1}{2} 	imes 6 	imes 8 = 24 \, cm^2$. Alternatively,using $AC$ as the base and $BM$ as the altitude,the area is $\frac{1}{2} 	imes AC 	imes BM = \frac{1}{2} 	imes 10 	imes BM = 5 	imes BM$. Equating the two areas: $5 	imes BM = 24$. Therefore,$BM = \frac{24}{5} = 4.8 \, cm$.

In $\Delta ABC$,$AB = 6 \, cm$,$BC = 8 \, cm$,and $AC = 10 \, cm$. $BM$ is an altitude on $AC$. Then $BM = \dots \, cm$.

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