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

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Question

(B) In a right-angled triangle,if an altitude is drawn from the right-angle vertex to the hypotenuse,the altitude is the geometric mean of the segment

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) In a right-angled triangle,if an altitude is drawn from the right-angle vertex to the hypotenuse,the altitude is the geometric mean of the segments of the hypotenuse.According to the geometric mean theorem: $BM^2 = AM \cdot CM$.Given $AM = 2x$,$BM = 3x + 5$,and $CM = 8x$.Substituting these values into the equation: $(3x + 5)^2 = (2x)(8x)$.Expanding the left side: $9x^2 + 30x + 25 = 16x^2$.Rearranging the terms to form a quadratic equation: $7x^2 - 30x - 25 = 0$.Factoring the quadratic equation: $7x^2 - 35x + 5x - 25 = 0$.$7x(x - 5) + 5(x - 5) = 0$.$(7x + 5)(x - 5) = 0$.This gives $x = 5$ or $x = -5/7$.Since $x$ represents a length,it must be positive,so $x = 5$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 2x$,$BM = 3x + 5$,and $CM = 8x$,find the value of $x$.

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