In Delta ABC,m angle B = 90^{circ} and overli | vedclass

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(B) In $\Delta ABC$,$m \angle B = 90^{\circ}$.Using the Pythagorean theorem,$AC^2 = AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100$.Therefore,$AC = \sqrt{100

Vedclass · Accepted Answer

$4.8$

Vedclass · Answer

(B) In $\Delta ABC$,$m \angle B = 90^{\circ}$.Using the Pythagorean theorem,$AC^2 = AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100$.Therefore,$AC = \sqrt{100} = 10$.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$.Equating the two expressions: $\frac{1}{2} 	imes AB 	imes BC = \frac{1}{2} 	imes AC 	imes BM$.$AB 	imes BC = AC 	imes BM$.$8 	imes 6 = 10 	imes BM$.$48 = 10 	imes BM$.$BM = \frac{48}{10} = 4.8$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude. If $AB = 8$ and $BC = 6$,then $BM = \dots$

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