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

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(A) Let $AM = x$. Since $AC = 25$,then $MC = 25 - x$.In $\Delta ABC$,$\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$. By the Geometr

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$AB = 20, BC = 15$

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(A) Let $AM = x$. Since $AC = 25$,then $MC = 25 - x$.In $\Delta ABC$,$\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$. By the Geometric Mean Theorem,$BM^2 = AM \cdot MC$.Substituting the values: $12^2 = x(25 - x) \implies 144 = 25x - x^2$.Rearranging gives the quadratic equation: $x^2 - 25x + 144 = 0$.Factoring the equation: $(x - 16)(x - 9) = 0$,so $x = 16$ or $x = 9$.Given $AM > CM$,we choose $AM = 16$ and $MC = 9$.Using the Pythagorean theorem in $\Delta ABM$: $AB^2 = AM^2 + BM^2 = 16^2 + 12^2 = 256 + 144 = 400 \implies AB = 20$.Using the Pythagorean theorem in $\Delta CBM$: $BC^2 = MC^2 + BM^2 = 9^2 + 12^2 = 81 + 144 = 225 \implies BC = 15$.Thus,$AB = 20$ and $BC = 15$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$,$\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$,and $AM > CM$. If $BM = 12$ and $AC = 25$,find $AB$ and $BC$.

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