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

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(A) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$,we use the geometric mean theorem (o

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$12$

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(A) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$,we use the geometric mean theorem (or properties of similar triangles).According to the property of altitudes in a right triangle,the square of the leg is equal to the product of the hypotenuse segment adjacent to it and the entire hypotenuse.Specifically,$BC^2 = CM 	imes AC$.Given $AM = 7$ and $CM = 9$,the total length of the hypotenuse $AC = AM + CM = 7 + 9 = 16$.Substituting the values into the formula: $BC^2 = 9 	imes 16$.$BC^2 = 144$.Taking the square root of both sides,$BC = \sqrt{144} = 12$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$. If $AM = 7$ and $CM = 9$,then $BC = \ldots$.

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