In Delta ABC,mangle B = 90^{circ} and overlin | vedclass

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(A) In a right-angled triangle,the altitude to the hypotenuse divides the triangle into two triangles similar to the original triangle and to each oth

Vedclass · Accepted Answer

$BM = 4x^2, AB = 2\sqrt{5}x^2, BC = 4\sqrt{5}x^2$

Vedclass · Answer

(A) In a right-angled triangle,the altitude to the hypotenuse divides the triangle into two triangles similar to the original triangle and to each other.By the geometric mean theorem,$BM^2 = AM \cdot CM$.Substituting the given values: $BM^2 = (2x^2)(8x^2) = 16x^4$.Taking the square root: $BM = 4x^2$.Using the Pythagorean theorem in $\Delta ABM$: $AB^2 = AM^2 + BM^2 = (2x^2)^2 + (4x^2)^2 = 4x^4 + 16x^4 = 20x^4$.Thus,$AB = \sqrt{20x^4} = 2\sqrt{5}x^2$.Using the Pythagorean theorem in $\Delta BCM$: $BC^2 = CM^2 + BM^2 = (8x^2)^2 + (4x^2)^2 = 64x^4 + 16x^4 = 80x^4$.Thus,$BC = \sqrt{80x^4} = 4\sqrt{5}x^2$.

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

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