1. Problem statement: Find the perimeter of triangle ABC whose vertices are A(2,1), B(5,1), and C(5,5). 2. Formula and rules: We use the distance formula to compute each side length. $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ Important rule: If two points share the same x or y coordinate, the distance equals the absolute difference of the other coordinate, which simplifies calculations. 3. Compute $AB$: Points A(2,1) and B(5,1) share the y-coordinate, so $$AB=\sqrt{(5-2)^2+(1-1)^2}=\sqrt{3^2+0^2}=\sqrt{9}=3$$ 4. Compute $BC$: Points B(5,1) and C(5,5) share the x-coordinate, so $$BC=\sqrt{(5-5)^2+(5-1)^2}=\sqrt{0^2+4^2}=\sqrt{16}=4$$ 5. Compute $AC$: Points A(2,1) and C(5,5) give $$AC=\sqrt{(5-2)^2+(5-1)^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$$ 6. Perimeter: Sum the side lengths to get the perimeter $P$. $$P=AB+BC+AC=3+4+5=12$$ 7. Final answer: The perimeter of triangle ABC is 12.