There is a point $P$ inside equilateral $\triangle ABC$ such that $PA=PB=PC=x.$ We know $[ABC]=\dfrac{\sqrt 3}{3}.$ Find $x.$

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There is a point $P$ inside equilateral $\triangle ABC$ such that $PA=PB=PC=x.$ We know $[ABC]=\dfrac{\sqrt 3}{3}.$ Find $x.$

CPhill · Answer 1 · 2024-07-20T10:32:10

sqrt (3) / 3 = (1/2) s^2 * sqrt (3) / 2 where s is the side of the triangle

1/3 = (1/4)s^2

4/3 = s^2

We have triangle APB

angle PAB = angle PBA = 30

angle APB = 120

Law of Cosines

s^2 = 2x^2 - 2x^2 * cos (120)

s^2 = 2x^2 - 2x^2 *(-1/2)

s^2 = 3x^2

4/3 = 3x^2

4/9 = x^2

2/3 = x

There is a point PP inside equilateral △ABC\triangle ABC such that PA=PB=PC=x.PA=PB=PC=x. We know [ABC]=√33.[ABC]=\dfrac{\sqrt 3}{3}. Find x.x.