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

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(C) In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$.By the properties of right-angled t

Vedclass · Accepted Answer

$25$

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(C) In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$.By the properties of right-angled triangles with an altitude to the hypotenuse,we have:$AB^{2} = AM \cdot AC$ and $BC^{2} = MC \cdot AC$Subtracting these two equations:$AB^{2} - BC^{2} = AM \cdot AC - MC \cdot AC$$AB^{2} - BC^{2} = AC(AM - MC)$Given that $AB^{2} - BC^{2} = 175$ and $AM - MC = 7$,we substitute these values:$175 = AC(7)$$AC = \frac{175}{7}$$AC = 25$

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM - MC = 7$ and $AB^{2} - BC^{2} = 175$,then $AC = \ldots$

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