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

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(B) In a right-angled triangle $\Delta ABC$ with altitude $\overline{AM}$ to the hypotenuse $\overline{BC}$,we have the geometric mean theorem: $AM^2

Vedclass · Accepted Answer

$12 + 4\sqrt{3}$

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(B) In a right-angled triangle $\Delta ABC$ with altitude $\overline{AM}$ to the hypotenuse $\overline{BC}$,we have the geometric mean theorem: $AM^2 = BM \cdot CM$.Given $BM = 6$ and $CM = 2$,$AM^2 = 6 \cdot 2 = 12$,so $AM = \sqrt{12} = 2\sqrt{3}$.Now,calculate the sides of $\Delta ABC$:$BC = BM + CM = 6 + 2 = 8$.Using the Pythagorean theorem in $\Delta ABM$: $AB^2 = AM^2 + BM^2 = 12 + 36 = 48$,so $AB = \sqrt{48} = 4\sqrt{3}$.Using the Pythagorean theorem in $\Delta ACM$: $AC^2 = AM^2 + CM^2 = 12 + 4 = 16$,so $AC = \sqrt{16} = 4$.Perimeter of $\Delta ABC = AB + AC + BC = 4\sqrt{3} + 4 + 8 = 12 + 4\sqrt{3}$.

In $\Delta ABC$,$m \angle A = 90^{\circ}$ and $\overline{AM}$ is an altitude to the hypotenuse $\overline{BC}$. If $BM = 6$ and $CM = 2$,find the perimeter of $\Delta ABC$.

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