Question

Let in a $△ABC,$ $x,y,z$ are the lengths of perpendicular drawn from the vertices of the triangle to the opposite sides $a,b,c$ and $\frac{bx}{c}+\frac{cy}{a}+\frac{az}{b}=\frac{a^2+b^2+c^2}{k},$ then the value of $k$ is
(where $R$ and $S$ are circumradius and semiperimeter respectively.)

• A. $R$
• B. $S$
• C. $2R$
• D. $2S$

Answer 1

The correct option is C $2R$

We know perpendicular distances :
$x=c\sin B,y=a\sin C,z=b\sin A$
Then,
$\frac{bx}{c}+\frac{cy}{a}+\frac{az}{b}=b\sin B+c\sin C+a\sin A$
Also, $\sin A=\frac{a}{2R},\sin B=\frac{b}{2R},\sin C=\frac{c}{2R}$
put in above equation :
$\frac{bx}{c}+\frac{cy}{a}+\frac{az}{b}=\frac{a^2+b^2+c^2}{2R}=\frac{a^2+b^2+c^2}{k}\therefore k=2R$