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

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Question

(A) In a right-angled triangle $\Delta ABC$ with altitude $\overline{BM}$ to the hypotenuse $\overline{AC}$,we have the geometric mean property: $BM^2

Vedclass · Accepted Answer

$2 \sqrt{13}$

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(A) In a right-angled triangle $\Delta ABC$ with altitude $\overline{BM}$ to the hypotenuse $\overline{AC}$,we have the geometric mean property: $BM^2 = AM \cdot CM$.Given $BM = 6$,we have $AM \cdot CM = 6^2 = 36$.We also know $AM + CM = AC = 13$.Let $AM = x$ and $CM = y$. Then $x + y = 13$ and $xy = 36$.These are the roots of the quadratic equation $t^2 - 13t + 36 = 0$.Factoring the equation: $(t - 9)(t - 4) = 0$,so $t = 9$ or $t = 4$.Since $AM In $\Delta ABM$,by the Pythagorean theorem: $AB^2 = AM^2 + BM^2$.$AB^2 = 4^2 + 6^2 = 16 + 36 = 52$.$AB = \sqrt{52} = \sqrt{4 \cdot 13} = 2 \sqrt{13}$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$,$\overline{BM}$ is an altitude to the hypotenuse $AC$,and $AM < CM$. If $BM = 6$ and $AC = 13$,find $AB$.

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