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

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(A) In a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other.By the geometri

Vedclass · Accepted Answer

$AB = 6, BC = 3\sqrt{5}, BM = 2\sqrt{5}$

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(A) In a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other.By the geometric mean theorem for the altitude: $BM^2 = AM \cdot CM$.Substituting the given values: $BM^2 = 4 \cdot 5 = 20$.Therefore,$BM = \sqrt{20} = 2\sqrt{5}$.Using the Pythagorean theorem in $\Delta ABM$: $AB^2 = AM^2 + BM^2 = 4^2 + (2\sqrt{5})^2 = 16 + 20 = 36$.Therefore,$AB = \sqrt{36} = 6$.Using the Pythagorean theorem in $\Delta CBM$: $BC^2 = CM^2 + BM^2 = 5^2 + (2\sqrt{5})^2 = 25 + 20 = 45$.Therefore,$BC = \sqrt{45} = 3\sqrt{5}$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 4$ and $CM = 5$,find $AB$,$BC$,and $BM$.

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