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

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(D) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,if $\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$,then by

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$5$

Vedclass · Answer

(D) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$,if $\overline{BM}$ is the altitude to the hypotenuse $\overline{AC}$,then by the geometric mean theorem,$BM^2 = AM \cdot CM$.Substituting the given values: $(x + 1)^2 = (x - 1)(x + 4)$.Expanding both sides: $x^2 + 2x + 1 = x^2 + 4x - x - 4$.Simplifying the equation: $x^2 + 2x + 1 = x^2 + 3x - 4$.Subtracting $x^2$ from both sides: $2x + 1 = 3x - 4$.Rearranging the terms: $3x - 2x = 1 + 4$.Thus,$x = 5$.

In $\Delta ABC, m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = x - 1$,$BM = x + 1$,and $CM = x + 4$,find the value of $x$.

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