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

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Question

(B) 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

$6$

Vedclass · Answer

(B) 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,the ratio of their corresponding sides is equal:$\frac{AM}{BM} = \frac{BM}{CM}$.Substituting the given values:$\frac{12}{BM} = \frac{BM}{3}$.Cross-multiplying gives:$BM^2 = 12 	imes 3$.$BM^2 = 36$.Taking the square root of both sides:$BM = \sqrt{36} = 6$.Therefore,the length of $BM$ is $6$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AM = 12$ and $CM = 3$,then $BM = \dots$

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