To solve the problem of finding the area of an equilateral triangle (ABC) where a point (P) inside it satisfies (PA=3), (PB=4), (PC=5), we use the rotation method:

Step 1: Rotation of Triangle (APB)

Rotate (\triangle APB) around (B) by (60^\circ) to get (\triangle CP'B). This gives:

• (BP=BP'=4) (rotation preserves length),
• (\angle PBP'=60^\circ) (rotation angle), so (\triangle PBP') is equilateral ((PP'=4)),
• (P'C=PA=3) (rotation preserves length).

Step 2: Identify Right Triangle (PP'C)

Check (\triangle PP'C):
(PP'=4), (P'C=3), (PC=5).
Since (3^2 + 4^2 = 5^2), (\triangle PP'C) is a right triangle ((\angle PP'C=90^\circ)).

Step 3: Calculate (\angle BP'C)

(\angle BP'C = \angle BP'P + \angle PP'C = 60^\circ + 90^\circ = 150^\circ).

Step 4: Compute (BC^2) Using Law of Cosines

For (\triangle BP'C):
(BC^2 = BP'^2 + P'C^2 - 2 \cdot BP' \cdot P'C \cdot \cos(150^\circ))
(\cos(150^\circ)=-\frac{\sqrt{3}}{2}), so:
(BC^2 = 4^2 + 3^2 - 2 \cdot 4 \cdot 3 \cdot \left(-\frac{\sqrt{3}}{2}\right) = 25 + 12\sqrt{3}).

Step 5: Area of Equilateral Triangle (ABC)

Area of equilateral triangle is (\frac{\sqrt{3}}{4} \cdot BC^2):
[Area = \frac{\sqrt{3}}{4}(25 + 12\sqrt{3}) = \frac{25\sqrt{3}}{4} + 9 = \frac{25\sqrt{3} + 36}{4}]

Answer: (\boxed{\frac{25\sqrt{3} + 36}{4}}) (or approximately (\boxed{20}) if rounded, but exact form is preferred).

(\boxed{\dfrac{25\sqrt{3} + 36}{4}})

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