In Delta ABC,angle B = 90^{circ} and BM is an | vedclass

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(A) In $\Delta ABC$,since $\angle B = 90^{\circ}$,by the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.Substituting the values,$AC^2 = 12^2 + 16^2 = 144 +

Vedclass · Accepted Answer

$9.6$

Vedclass · Answer

(A) In $\Delta ABC$,since $\angle B = 90^{\circ}$,by the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.Substituting the values,$AC^2 = 12^2 + 16^2 = 144 + 256 = 400$.Thus,$AC = \sqrt{400} = 20 \, cm$.The area of $\Delta ABC$ can be calculated as $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Using $AB$ and $BC$ as base and height,Area $= \frac{1}{2} 	imes 12 	imes 16 = 96 \, cm^2$.Alternatively,using $AC$ as the base and $BM$ as the altitude,Area $= \frac{1}{2} 	imes AC 	imes BM$.Equating the two expressions for area: $\frac{1}{2} 	imes 20 	imes BM = 96$.$10 	imes BM = 96$.$BM = 9.6 \, cm$.

In $\Delta ABC$,$\angle B = 90^{\circ}$ and $BM$ is an altitude to the hypotenuse $AC$. If $AB = 12 \, cm$ and $BC = 16 \, cm$,then find the length of $BM$ in $cm$.

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