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

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(D) 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

$4$

Vedclass · Answer

(D) 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 similarity of these triangles,we have the ratio of corresponding sides:$\frac{AM}{BM} = \frac{BM}{CM}$.This leads to the geometric mean theorem: $BM^2 = AM 	imes CM$.Given $AM = 8$ and $CM = 2$,we substitute these values into the equation:$BM^2 = 8 	imes 2 = 16$.Taking the square root of both sides,we get $BM = \sqrt{16} = 4$.Therefore,the length of $BM$ is $4$.

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

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