Question

$\Delta$ABC is an equilateral triangle of side 2$\sqrt{3}$ cm. P is any point in the interior of $\Delta$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)

$3cms$

Given, $a=2\sqrt{3}$
$x,y,z$ are perpendicular distances from the point P.
Area of triangle $=\frac{1}{2}\times base\times height$
Area of equilateral triangle $=\frac{\sqrt{3}}{4}{a}^2$
Area of $△ABC$ $=$ Sum of areas of smaller triangle whose perpendiculars are $x,y,z$
$\frac{1}{2}a(x+y+z)=\frac{\sqrt{3}}{4}{a}^2$
$x+y+z=\frac{\sqrt{3}}{2}a=\frac{\sqrt{3}}{2}\times 2\sqrt{3}==3$ cm
Hence, option 'C' is correct.