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

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(A) In a right-angled triangle $\Delta ABC$ with altitude $\overline{BM}$ to the hypotenuse $\overline{AC}$,the geometric mean theorem states that $BM

Vedclass · Accepted Answer

$4$

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(A) In a right-angled triangle $\Delta ABC$ with altitude $\overline{BM}$ to the hypotenuse $\overline{AC}$,the geometric mean theorem states that $BM^2 = AM \cdot CM$.Given $AM = x + 7$,$BM = x + 2$,and $CM = x$.Substituting these values into the formula: $(x + 2)^2 = (x + 7)(x)$.Expanding both sides: $x^2 + 4x + 4 = x^2 + 7x$.Subtracting $x^2$ from both sides: $4x + 4 = 7x$.Rearranging the terms: $4 = 7x - 4x$,which simplifies to $4 = 3x$.Therefore,$x = \frac{4}{3}$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude. If $AM = x + 7$,$BM = x + 2$,and $CM = x$,find the value of $x$.

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