Question

Let the point P lies in the interior of an equilateral triangle with side lengths 2 units, each. Find
the sum of the shortest distances from P to the sides of the triangle.


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Answer 1

Answer: √3 units

Step-by-step explanation:

1) Let ABC be an equilateral triangle with P be any interior point in it.

Smallest distance from P to its sides will be perpendicular distance from P to the sides which intersects AB ,BC and AC at D, E ,and F respectively.

2) Since,  all sides are equal

=> AB= BC = AC = 2=a (say)

Area of triangle :

$ar(\triangle ABC) =ar(\triangle APB) +ar(\triangle BPC) +ar(\triangle APC) \\ \\=>\frac{\sqrt{3}}{4}a^2=\frac{1}{2}*AB*PD+\frac{1}{2}*BC*PE+\frac{1}{2}*AC*PF \\ \\=> \frac{\sqrt{3}}{4}*2^2=\frac{1}{2}*a*(PD+PE+PF) \\ \\=>(PD+PE+PF)=\sqrt{3}$

Hence,  sum of shortest distance from P to sides of triangle is √3 units.

Answer 2

Answer:

answer is above

Step-by-step explanation:

answer is correct don't hesitate

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