In Delta ABC,mangle B = 90^{circ} and overlin | 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 to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) In a right-angled triangle,the altitude 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 $\Delta AMB \sim \Delta ABC$,we have the property: $BM^2 = AM \cdot MC$.Given $AC = 25$ and $AM = 16$,we find $MC = AC - AM = 25 - 16 = 9$.Substituting these values into the formula: $BM^2 = 16 \cdot 9$.$BM^2 = 144$.Taking the square root,$BM = \sqrt{144} = 12$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AC = 25$ and $AM = 16$,then $BM = \dots$

Similar Questions

In $\Delta ABC$ and $\Delta XYZ$,$\angle A \cong \angle Y$ and $\angle B \cong \angle Z$. If $\frac{AC}{YX} = \frac{5}{7}$ and $AB = 7$,find $YZ$.

In the figure,$ABC$ and $DBC$ are two triangles on the same base $BC$. If $AD$ intersects $BC$ at $O$,show that $\frac{\operatorname{ar}(ABC)}{\operatorname{ar}(DBC)} = \frac{AO}{DO}$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is a median. If $m \angle C = 30^{\circ}$ and $AB = 5$,find $BD$.

In the figure,$ABD$ is a triangle right-angled at $A$ and $AC \perp BD$. Show that $AC^{2} = BC \cdot DC$.

In the figure,$\Delta ODC \sim \Delta OBA$,$\angle BOC = 125^{\circ}$ and $\angle CDO = 70^{\circ}$. Find $\angle DOC$,$\angle DCO$,and $\angle OAB$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module