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

(B) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$,the triangle is divided into two tri

Vedclass · Accepted Answer

$AM$ and $AC$

Vedclass · Answer

(B) In a right-angled triangle $\Delta ABC$ where $\angle B = 90^{\circ}$ and $\overline{BM} \perp \overline{AC}$,the triangle is divided into two triangles $\Delta AMB$ and $\Delta BMC$ which are similar to the original triangle $\Delta ABC$ and to each other.By the property of similarity,$\Delta AMB \sim \Delta ABC$.This implies that the ratio of corresponding sides is equal: $\frac{AM}{AB} = \frac{AB}{AC}$.Cross-multiplying gives $AB^2 = AM \cdot AC$.Therefore,$AB$ is the geometric mean of $AM$ and $AC$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $AC$. Then,$AB$ is the geometric mean of.......

Similar Questions

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $AB = 20$ and $AC = 29$,find the perimeter of $\Delta ABC$.

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. If $AB = 4, BC = 8, AC = 10$ and $PR = 15$,then the perimeter of $\Delta PQR = \dots$

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 5$,$MB = 8$ and $AN = 10$,then $NC = \dots$

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BE}$ is a median. If $AB = 15$ and $BE = 8.5$,find $BC$.

If $\Delta ABC \sim \Delta DEF$ for the correspondence $ABC \leftrightarrow DEF$,and given $AB = 8$,$AC = 10$,$DE = 12$,and $EF = 18$,find the perimeter of $\Delta DEF$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module