Tangle with triangle

Area of triangle $ABC$ where $u,v,w$ are its medians is given by :

$\dfrac{1}{3}\sqrt{[2(u^2v^2+v^2w^2+u^2w^2)-(u^4+v^4+w^4)]}$

Substituting the values of medians , we get

$A(ABC)=6$

Since , one of vertices of triangles is refected , it can be proved easily by congruence that :

$A(ABC)=A(BPC)=6$

$\Rightarrow A(ABCP)=A(ABC)+A(BPC)=\huge\boxed{\color{#3D99F6}{12}}$

I used coordinate geometry

Let A(0,a) B(-b,0) & C(c,0)

Then E((c-b)/2,0) D(-b/2,a/2) & F(c/2,a/2)

Using diatance formula we get following equations :

b²+c²-2bc+4a²=25....(1)

4b²+c²+4bc+a²=73....(2)

b²+4c²+4bc+a²=52....(3)

From (2) & (3),

b²-c²=7 &

5b²+5c²+8bc+2a²=125

5(b²+c²-2bc)+18bc+2a²=125

5(25-4a²)+18bc+2a²=125

a²=bc

Plugging this in (1) we get (b+c)²=25 or b+c=5

Hence b-c=7/5

b=16/5, c=9/5 & a=12/5

Area of ABC is 1/2 x (b+c) x a= 6

Area of quad ABPC is 2ABC=12

There is another formula which can be applied to find out the area of a triangle. Let A be = (m1 + m2 + m3)/2, where m1, m2, m3 are the medians of the triangle. Then area of the triangle is given by the following expression:

4/3 x [(A)(A - m1)(A - m2)(A - m3)]^1/2

Sorry guys, still did not learn LaTex :(