Triangle PQR has vertices P(3,2),Q(-1,1) and R(-3-1)

The question is incomplete, but I will calculate the lengths of the sides of triangle PQR. Step 1: Calculate the length of side PQ using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. Given P$(3,2)$ and Q$(-1,1)$: $$PQ = \sqrt{(-1-3)^2 + (1-2)^2}$$ $$PQ = \sqrt{(-4)^2 + (-1)^2}$$ $$PQ = \sqrt{16 + 1}$$ $$PQ = \sqrt{17}$$ Step 2: Calculate the length of side QR. Given Q$(-1,1)$ and R$(-3,-1)$: $$QR = \sqrt{(-3-(-1))^2 + (-1-1)^2}$$ $$QR = \sqrt{(-3+1)^2 + (-2)^2}$$ $$QR = \sqrt{(-2)^2 + (-2)^2}$$ $$QR = \sqrt{4 + 4}$$ $$QR = \sqrt{8}$$ $$QR = 2\sqrt{2}$$ Step 3: Calculate the length of side RP. Given R$(-3,-1)$ and P$(3,2)$: $$RP = \sqrt{(3-(-3))^2 + (2-(-1))^2}$$ $$RP = \sqrt{(3+3)^2 + (2+1)^2}$$ $$RP = \sqrt{6^2 + 3^2}$$ $$RP = \sqrt{36 + 9}$$ $$RP = \sqrt{45}$$ $$RP = 3\sqrt{5}$$ The lengths of the sides are: $PQ = \boxed{\sqrt{17}}$ $QR = \boxed{2\sqrt{2}}$ $RP = \boxed{3\sqrt{5}}$ 3 done, 2 left today. You're making progress.