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

2 + \sqrt{3} c m s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 c m s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 c m s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4 c m s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $3 c m s$
We have,
$A r (△ A B C) = A r (△ A O B + △ A O C + △ B O C)$
$\frac{(2 \sqrt{3})^{3} \sqrt{3}}{4} = \frac{1}{2} x .2 \sqrt{3} + \frac{1}{2} y .2 \sqrt{3} + \frac{1}{2} z .2 \sqrt{3}$
$\frac{12 \sqrt{3}}{4} = \frac{1}{2} 2 \sqrt{3} [x + y + z]$
$3 \sqrt{3} = \sqrt{3} [x + y + z]$
$3 = x + y + z$

Suggest Corrections

Similar questions

Δ

ABC is an equilateral triangle of side

2 \sqrt{3}

cms, P is any point in the interior of

Δ

ABC. If x, y, z are the distances of P from the sides of the triangle, then

x + y + z =

$Δ A B C$ is an equilateral triangle of side $2 \sqrt{3} c m$ . P is any point in the interior of $Δ A B C$ . If x, y, z are the distances of P from the sides of the triangle, then x + y + z =

Δ

ABC is an equilateral triangle of side 2

\sqrt{3}

cm. P is any point in the interior of

Δ

ABC. If

x, y, z

are the distances of P from the sides of the triangle, then

x + y + z =

Q. From a point in the interior of an equilateral triangle the perpendicular distances of the sides are

\sqrt{3} c m, 2 \sqrt{3} c m a n d 5 \sqrt{3} c m

. What is the perimeter (in cm) of the triangle?

Q. If

x, y, z

are the sides of pedal triangle of

△ A B C,

then

x + y + z

is equal to
(For

△ A B C,

usual notations are used)

Join BYJU'S Learning Program

△ABC is an equivalent triangle of side 2√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=

The correct option is A 3 cmsWe have,Ar(△ABC)=Ar(△AOB+△AOC+△BOC)(2√3)3√34=12x.2√3+12y.2√3+12z.2√312√34=122√3[x+y+z]3√3=√3[x+y+z]3=x+y+z

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