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

(A) In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$.According to the geometric mean theorem for rig

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$.According to the geometric mean theorem for right-angled triangles,the altitude to the hypotenuse divides the triangle into two triangles similar to the original and to each other.Thus,$BM^2 = AM \cdot CM$.Given $AC = 13$ and $CM = 9$,we have $AM = AC - CM = 13 - 9 = 4$.Substituting these values into the formula: $BM^2 = 4 \cdot 9 = 36$.Therefore,$BM = \sqrt{36} = 6$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude. If $AC = 13$ and $CM = 9$,then $BM = \ldots \ldots$

Similar Questions

In the figure,two chords $AB$ and $CD$ of a circle intersect each other at point $P$ (when produced) outside the circle. Prove that $\Delta PAC \sim \Delta PDB$.

If $\triangle ABC \sim \triangle EDF$ and $\triangle ABC$ is not similar to $\triangle DEF$,then which of the following is not true?

$\Delta PQR \sim \Delta DEF$ for the correspondence $PQR \leftrightarrow DEF$. If $3 PQ = 2 DE$,$EF = 6$,and $PR = 8$,find $QR$ and $DF$.

$\Delta ABC \sim \Delta XYZ$ for the correspondence $ABC \leftrightarrow XYZ$. If $AB + BC = 7$,$XY + YZ = 10.5$,and $AC = 5$,then $XZ = \ldots$

In the figure,if $\Delta ABE \cong \Delta ACD,$ show that $\Delta ADE \sim \Delta ABC.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module