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

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(C) In a right-angled triangle,the altitude drawn from the right-angle vertex to the hypotenuse divides the triangle into two triangles that are simil

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$6$

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(C) In a right-angled triangle,the altitude drawn from the right-angle vertex to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.According to the geometric mean theorem,the altitude $\overline{BM}$ to the hypotenuse $\overline{AC}$ satisfies the relation: $BM^2 = AM 	imes CM$.Given $AM = 12$ and $CM = 3$.Substituting the values: $BM^2 = 12 	imes 3 = 36$.Taking the square root of both sides: $BM = \sqrt{36} = 6$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 12$ and $CM = 3$,find $BM$.

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