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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $BM = \dots$

Similar Questions

In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. If $MA = 10$,$MB = 8$,and $MC = 5$,find $BD$.

In the figure,if $\angle ACB = \angle CDA$,$AC = 8 \, cm$ and $AD = 3 \, cm$,find $BD$ (in $cm$).

In an isosceles right-angled triangle, the length of the hypotenuse is $24$. Find the area of the triangle.

In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. If $\frac{AB}{CD} = \frac{2}{1}$ and $AC = 15$,find $MA$ and $MC$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. Then the correspondence $ADB \leftrightarrow \ldots$ between $\Delta ADB$ and $\Delta BDC$ is a similarity.

Vedclass Test Series

Exam Paper Generator

Online Exam Module