Finding the area of the square?

I got 232 for the area of ABCD.

Rotate the triangle ABP at a right angle clockwise around B.to position CBP'. Rotate the triangle CDP at a right angle clockwise around D.to position ADP". Now the area of the square is equal to the sum of the areas of the quadrilaterals BPCP' (with a right angle at B) and APDP" (with a right angle at D). Next:

Area(BPCP') = Area(BPP') + Area(CPP');

Area(APDP") = Area(APP") + Area(DPP");

Area(BPP') = 6*6/2 = 18, sides 6, 6, 6√2;

Area(CPP') = 42 by Heron's formula, sides 10, 14, 6√2;

Area(APP") =130, sides 2√65, 2√65, 2√130;

Area(DPP") = 42 by Heron's formula, sides 10, 14, 2√130.

Finally Area(ABCD) = 18 + 2*42 + 130 = 232

I'd welcome shorter solutions if any.

One way is to use the Law of Cosines to find the side length.

Letting s = length of the side,

You can apply Law of Cosines to

ABP (using the angle at B) and BCP (also using the angle at B, which is the complement of the angle at B in ABP).

For ABP:

14^2 = s^2 + 6^2 - 2 (6) s Cos(theta)

For BCP

10^2 = s^2 + 6^2 - 2 (6) s Sin(theta)

Thus cos theta comes up in one equation, sin theta in the other because the relevant angles are complements. You can eliminate the angle theta using the identity Cos^2 theta + Sin^2 theta = 1, leaving an equation in s alone. You can solve for s, getting two possibilities, 8 and 2 Sqrt(58). Probably using a second pair of adjacent triangles in the same way eliminates the ambiguity. You can try it out anyway. The area is either 8^2 = 64 or (2 Sqrt(58))^2 = 232.

Here's how I would do it:

Just start making triangles with point p in the middle.

Triangle APB

Triangle BPC

Triangle CPD

Triangle DPA

Once you have those, technically, you know that all of the sides are equal, so AB=BC=CD=DA.

The basic way to find area is to multiply two of those sides together. Which equals the area of all of the triangles, so you only have one variable if you set them equal to one another, the lenght of one side.

Give me a couple minutes to write it all out on paper, but I'll add my solution in a bit.