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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) In a right-angled triangle $\Delta ABC$ with $\angle B = 90^\circ$,if $\overline{BM}$ is the altitude to the hypotenuse $AC$,then by the Geometric Mean Theorem (or properties of similar triangles $\Delta BMC \sim \Delta BCA$),we have the relation: $BC^2 = CM \cdot AC$.Given $BC = 2x + 2$,$CM = x + 1$,and $AC = 5x$.Substituting these values into the equation: $(2x + 2)^2 = (x + 1)(5x)$.Expanding the left side: $(2(x + 1))^2 = 4(x + 1)^2 = 4(x^2 + 2x + 1)$.So,$4(x^2 + 2x + 1) = 5x^2 + 5x$.$4x^2 + 8x + 4 = 5x^2 + 5x$.Rearranging the terms to form a quadratic equation: $x^2 - 3x - 4 = 0$.Factoring the quadratic: $(x - 4)(x + 1) = 0$.This gives $x = 4$ or $x = -1$.Since $x$ represents a length component and $BC, CM, AC$ must be positive,we discard $x = -1$.Therefore,$x = 4$.

In $\Delta ABC$,$m \angle B = 90^\circ$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$. If $BC = 2x + 2$,$CM = x + 1$,and $AC = 5x$,find the value of $x$.

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