For this question, look for clues in the lengths of the sides as well as in the answer choices. Those \(\sqrt{2}\)'s suggest that special right triangles may be involved.

The pentagon can be broken up into two isosceles right triangles (45-45-90), one on the left side of the figure, one on the right. Because the legs of these triangles are each 2, the hypotenuse is \(2\sqrt{2}\).

That leaves one remaining triangle in the middle of the figure. This triangle must be equilateral, with sides of \(2\sqrt{2}\).

Since an equilateral triangle is just two 30-60-90 triangles put together, we can figure out the height of the triangle. Remember, the ratio for a
30-60-90 triangle is \(x:2x:x\sqrt{3}\)

So the height of the equilateral triangle is \(\sqrt{2}\sqrt{3}\) or \(\sqrt{6}\).

The total area is the sum of the two 45-45-90 triangles (1/2 base * height for both triangles gives us 4) plus the area of the equilateral triangle \(\frac{1}{2}*2\sqrt{2}\sqrt{6}=\sqrt{12}\)

Total area is \(4+\sqrt{12}=4+2\sqrt{3}\)

Answer: A