We can find the distances between consecutive points to find the lengths of the sides of the figure ABCD. The distance between A and B is $\sqrt{(2-0)^2 + (5-1)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$. The distance between B and C is $\sqrt{(5-2)^2 + (2-5)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$. The distance between C and D is $\sqrt{(7-5)^2 + (0-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$. The distance between D and A is $\sqrt{(0-7)^2 + (1-0)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$. 

The perimeter of the figure ABCD is $2\sqrt{5} + 3\sqrt{2} + 2\sqrt{2} + 5\sqrt{2} = 2\sqrt{5} + 10\sqrt{2}$. 

Thus, the sum of a and b is $2 + 10 = \boxed{12}$. The answer is: 12
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
The points A=(0,1), B=(2,5), C=(5,2), and D=(7,0) are connected to form a figure ABCD. The perimeter of ABCD can be represented as a√2 + b√5, where a and b are integers. Find the sum of a and b.

### Answer:
