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 $\cot A+\cot B+\cot C=k(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}),$ then the value of $k$ is
(where $R,r,S,\Delta$ are circumradius,inradius,semiperimeter and area of triangle respectively.)

• A. $R^2$
• B. $rR$
• C. $\Delta$
• D. $a^2+b^2+c^2$

Answer 1

The correct option is C $\Delta$

We know,
$\Delta =\frac{1}{2}\left(ax\right)=\frac{1}{2}\left(by\right)=\frac{1}{2}\left(cz\right)$
Now,
$(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})=\frac{a^2}{4{\Delta }^2}+\frac{b^2}{4{\Delta }^2}+\frac{c^2}{4{\Delta }^2}=\frac{(a^2+b^2+c^2)}{4{\Delta }^2}\cdots \left(i\right)$
Also,
$\cot A+\cot B+\cot C=\frac{R}{abc}(b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2)\left[\text{using cosine rule}\right]=\frac{R}{abc}(a^2+b^2+c^2)=\frac{4{\Delta }^{2}R}{abc}(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})\left[\text{using}\right(i\left)\right]=\frac{4\Delta R}{abc}\cdot \Delta (\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})=\Delta \cdot (\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})[\because \Delta =\frac{abc}{4R}]$
$\therefore k=\Delta$