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

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Question

(C) In a right-angled triangle,the altitude drawn to the hypotenuse divides the triangle into two triangles that are similar to the original triangle

Vedclass · Accepted Answer

$2\sqrt{6}$

Vedclass · Answer

(C) In a right-angled triangle,the altitude drawn to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.Specifically,$\Delta AMB \sim \Delta BMC$.From the property of similarity,the ratio of corresponding sides is equal:$\frac{AM}{BM} = \frac{BM}{CM}$.This implies $BM^2 = AM 	imes CM$.Given $AM = 4$ and $CM = 6$,we have:$BM^2 = 4 	imes 6 = 24$.Taking the square root of both sides:$BM = \sqrt{24} = \sqrt{4 	imes 6} = 2\sqrt{6}$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 4$ and $CM = 6$,then $BM = \dots$

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