In Delta ABC,mangle B = 90^{circ} and overlin | vedclass

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(C) In a right-angled triangle,the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) In a right-angled triangle,the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.Specifically,$\Delta AMB \sim \Delta BMC$.From the similarity $\Delta AMB \sim \Delta ABC$,we have the property: $BM^2 = AM \cdot MC$.Given $AC = 25$ and $AM = 16$,we find $MC = AC - AM = 25 - 16 = 9$.Substituting these values into the formula: $BM^2 = 16 \cdot 9$.$BM^2 = 144$.Taking the square root,$BM = \sqrt{144} = 12$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AC = 25$ and $AM = 16$,then $BM = \dots$

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