Question

$△ABC$ is an equivalent triangle of side $2\sqrt{3}$ cms. $P$ is any point in the interior of $△ABC$. If $x,y,z$ are the distances of $P$ from the sides of the triangle, then $x+y+z=$

• A. $2+\sqrt{3}cms$
• B. $5cms$
• C. $3cms$
• D. $4cms$

Answer 1

Answer: (C)

$3cms$

We have,
$Ar\left(△ABC\right)=Ar(△AOB+△AOC+△BOC)$
$\frac{(2\sqrt{3}{)}^3\sqrt{3}}{4}=\frac{1}{2}x.2\sqrt{3}+\frac{1}{2}y.2\sqrt{3}+\frac{1}{2}z.2\sqrt{3}$
$\frac{12\sqrt{3}}{4}=\frac{1}{2}2\sqrt{3}[x+y+z]$
$3\sqrt{3}=\sqrt{3}[x+y+z]$
$3=x+y+z$