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

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(C) 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

Vedclass · Accepted Answer

$10$

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(C) 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 (or property of similar triangles $\Delta AMB \sim \Delta ABC$),we have the relation: $AB^2 = AM \cdot AC$.Given values are $AB = x - 2$,$AM = x - 6$,and $AC = 2x - 4$.Substituting these into the equation: $(x - 2)^2 = (x - 6)(2x - 4)$.Expanding both sides: $x^2 - 4x + 4 = 2x^2 - 4x - 12x + 24$.Simplifying: $x^2 - 4x + 4 = 2x^2 - 16x + 24$.Rearranging the terms to form a quadratic equation: $x^2 - 12x + 20 = 0$.Factoring the quadratic equation: $(x - 10)(x - 2) = 0$.This gives $x = 10$ or $x = 2$.If $x = 2$,then $AB = 2 - 2 = 0$,which is impossible for a triangle side length. Thus,$x = 10$ is the only valid solution.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AB = x - 2$,$AM = x - 6$,and $AC = 2x - 4$,find the value of $x$.

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