Let $A=(0,1),$ $B=(2,5),$ $C=(5,2),$ and $D=(7,0).$ A figure is created by connecting $A$ to $B,$ $B$ to $C,$ $C$ to $D,$ and $D$ to $A.$ The perimeter of $ABCD$ can be expressed in the form $a\sqrt2+b\sqrt{5}$ with $a$ and $b$ integers. What is the sum of $a$ and $b$?
I need to find the lengths of each side of $ABCD$ and then add them up. To find the length of a side, I can use the distance formula, which is based on the Pythagorean theorem. The distance formula says that if I have two points $(x_1,y_1)$ and $(x_2,y_2),$ the distance between them is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$ So, for example, to find the length of $AB,$ I can plug in the coordinates of $A$ and $B$ into the distance formula: $\sqrt{(2-0)^2+(5-1)^2}=\sqrt{4+16}=\sqrt{20}.$ I can simplify $\sqrt{20}$ by noticing that it is equal to $\sqrt{4}\sqrt{5}=2\sqrt{5}.$ Similarly, I can find the lengths of the other sides of $ABCD$ by using the distance formula and simplifying the radicals. For $BC,$ I get $\sqrt{(5-2)^2+(2-5)^2}=\sqrt{9+9}=\sqrt{18}=\sqrt{9}\sqrt{2}=3\sqrt{2}.$ For $CD,$ I get $\sqrt{(7-5)^2+(0-2)^2}=\sqrt{4+4}=\sqrt{8}=\sqrt{4}\sqrt{2}=2\sqrt{2}.$ For $DA,$ I get $\sqrt{(0-7)^2+(1-0)^2}=\sqrt{49+1}=\sqrt{50}=\sqrt{25}\sqrt{2}=5\sqrt{2}.$ Now I have the lengths of all four sides of $ABCD:$ $2\sqrt{5},$ $3\sqrt{2},$ $2\sqrt{2},$ and $5\sqrt{2}.$ To find the perimeter, I just need to add them up: $2\sqrt{5}+3\sqrt{2}+2\sqrt{2}+5\sqrt{2}.$ I can group the terms with the same radical and combine their coefficients: $2\sqrt{5}+(3+2+5)\sqrt{2}.$ This simplifies to $2\sqrt{5}+10\sqrt{2}.$ Therefore, the perimeter of $ABCD$ is $2\sqrt{5}+10\sqrt{2}.$ The problem asks me to express the perimeter in the form $a\sqrt2+b\sqrt{5},$ where $a$ and $b$ are integers. Comparing this form with my answer, I can see that $a=10$ and $b=2.$ The sum of $a$ and $b$ is $10+2=12.$ 