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(A) In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$.According to the geometric mean theorem for rig

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$.According to the geometric mean theorem for right-angled triangles,the altitude to the hypotenuse divides the triangle into two triangles similar to the original and to each other.Thus,$BM^2 = AM \cdot CM$.Given $AC = 13$ and $CM = 9$,we have $AM = AC - CM = 13 - 9 = 4$.Substituting these values into the formula: $BM^2 = 4 \cdot 9 = 36$.Therefore,$BM = \sqrt{36} = 6$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude. If $AC = 13$ and $CM = 9$,then $BM = \ldots \ldots$

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