1. **State the problem:** We need to classify triangle $\triangle ABC$ with vertices $A(5,4)$, $B(2,8)$, and $C(6,5)$ based on its angle measures and side lengths. 2. **Find the lengths of the sides using the distance formula:** The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Calculate side $AB$:** $$AB = \sqrt{(2 - 5)^2 + (8 - 4)^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$ 4. **Calculate side $BC$:** $$BC = \sqrt{(6 - 2)^2 + (5 - 8)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$ 5. **Calculate side $AC$:** $$AC = \sqrt{(6 - 5)^2 + (5 - 4)^2} = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \approx 1.414$$ 6. **Classify by side lengths:** Since $AB = BC = 5$ and $AC \approx 1.414$, two sides are equal, so $\triangle ABC$ is isosceles. 7. **Check for right angle using the Pythagorean theorem:** Check if $AB^2 + AC^2 = BC^2$ or any permutation: $$AB^2 + AC^2 = 5^2 + (\sqrt{2})^2 = 25 + 2 = 27$$ $$BC^2 = 5^2 = 25$$ Not equal. Check $AB^2 + BC^2$: $$25 + 25 = 50$$ $$AC^2 = 2$$ Not equal. Check $AC^2 + BC^2$: $$2 + 25 = 27$$ $$AB^2 = 25$$ Not equal. No right angle. 8. **Check if triangle is acute or obtuse:** The largest side is $5$. Compare sum of squares of smaller sides to square of largest side: $$AC^2 + AC^2 = 2 + 25 = 27 > 25$$ Since sum of squares of smaller sides is greater than square of largest side, the triangle is acute. **Final classification:** $\triangle ABC$ is an **acute isosceles triangle**.