Let P be a point inside the equilateral triangle ABC, PA=3 ,PB=4,PC=5 then the area of the triangle ABC is what???

Question

ABC is a equilateral triangle therefore all the angles are 60° .And P is a point inside the triangle. And we have to find the area of the triangle.

Answer 1

Hint: Reflect $P$ in $AB$, $BC$ and $CA$ to obtain points $X$, $Y$ and $Z$ respectively. Then we have a hexagon doubling the area of the triangle. This hexagon can be cut into $\triangle XYZ$ and $3$ isosceles triangles. It's not difficult to find the areas of these $4$ triangles.

Answer 2

Start from an equilateral triangle with vertices $A, B, C$ ordered in a counterclockwise manner.

• Let $P$ be a point in its interior.
• Let $a, b, c$ be distances of $P$ to vertices $A, B, C$ respectively.
• Rotate $P$ with respect to the center of $\triangle ABC$ to point $Q$.
• Reflect $P$ with respect to midpoints of $AB$, $BC$ and $CA$ to obtain three points $C'$, $A'$ and $B'$.

In this way, we obtain three parallelograms $AC'BP$, $BA'CP$, $CB'AP$. The sum of their areas is the area of hexgaon $AC'BA'CB'$ which is twice of that for triangle $ABC$.

To compute the area of hexgaon $AC'BA'CB'$, we split it into $6$ triangles.

Let's look at one of the triangle, $\triangle QA'C$, first. Since $BA'CP$ is a parallelogram, $A'C = BP = b$. Since we obtain $Q$ by rotating $P$ for $120^\circ$, $CQ = BP = b$ too. Furthermore, since $A'C \parallel BP$, $\angle QCA' = 60^\circ$. This means $\triangle QAC'$ is an equilateral triangle and $QA' = b$.

By a similar argument, $\triangle AQB'$ is an equilateral triangle with side $a$. $\triangle C'BQ$ is an equilateral triangle with side $c$.

For the remaining three triangles $QBA'$, $QCB'$ and $QAC'$, it is easy to see they are congruent with sides $a, b, c$. Let $\Delta$ be the area of such a triangle.

Combine these, we obtain

$$2\verb/Area/(\triangle ABC) = \verb/Area/(AC'BA'CB') = \frac{\sqrt{3}}{4}(a^2+b^2+c^2) + 3\Delta$$

Apply this formula to the problem at hand where $a = 3, b = 4, c = 5$. Since a triangle with sides $3,4,5$ is a right angled triangle with area $\Delta = 6$, the desired area is

$$\frac12\left[\frac{\sqrt{3}}{4}(3^2 + 4^2 + 5^2) + 3\times 6\right] = 9 + \frac{25}{4}\sqrt{3}$$