Question

AD,BE,and CF are the perpendicular from the angular points of $\Delta ABC$ upon the opposite sides .The perimeters of the $\Delta DEF$ and $\Delta ABC$ are in the ratio :

• A. $\frac{2r}{R}$
• B. $\frac{r}{2R}$
• C. $\frac{r}{R}$
• D. $\frac{r}{3R}$

Answer 1

Answer: (C)

$\frac{r}{R}$

Given: $△ABC$ is $s.t.AD,BE,CF$ are perpendiculars from angular ots, to $△ABC$ on opposite sides

$\frac{Perimeterof△DEF}{Perimeterof△ABC}$

We know that $△DEF$ forms pedal triangle of

$\therefore \frac{Perimeterof△DEF}{Perimeterof△ABC}dfraca\cos A+b\cos B+c\cos Ca+b+c$

$=\frac{2R\sin A\cos A+2R\sin B\cos B+2R\sin C\times \sin C}{2R\sin A+2R\sin B+2R\sin C}$

By conditional Identities,

$=\frac{\sin \left(2A\right)+-\sin \left(2B\right)+\sin \left(2C\right)}{2(\sin A+\sin B+\sin C)}=\frac{4\sin A.\sin B.\sin C}{2(4\cos \frac{A}{2}.\cos \frac{B}{2}.\cos \frac{C}{2})}$

$=\frac{\left(2\sin \frac{A}{2}\cos \frac{A}{2}\right)\left(2\sin \frac{B}{2}\cos \frac{B}{2}\right)\left(2\sin \frac{C}{2}\cos \frac{C}{2}\right)}{\left(2\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}\right)}$

$=4\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}$

$\frac{Perimeterof△DEF}{Perimeterof△ABC}=\frac{r}{R}$

$C.\frac{r}{R}$