1. The problem asks which sets of three lengths can form a right triangle. 2. To determine if three lengths $a$, $b$, and $c$ form a right triangle, we use the Pythagorean theorem: $$a^2 + b^2 = c^2$$ where $c$ is the longest side (hypotenuse). 3. We check each set by identifying the longest side and verifying if the sum of the squares of the other two equals the square of the longest. 4. For (40, 50, 30): Longest side is 50. $$40^2 + 30^2 = 1600 + 900 = 2500$$ $$50^2 = 2500$$ Since $1600 + 900 = 2500$, this set forms a right triangle. 5. For (41, 9, 40): Longest side is 41. $$9^2 + 40^2 = 81 + 1600 = 1681$$ $$41^2 = 1681$$ Since $81 + 1600 = 1681$, this set forms a right triangle. 6. For (70, 20, 25): Longest side is 70. $$20^2 + 25^2 = 400 + 625 = 1025$$ $$70^2 = 4900$$ Since $1025 \neq 4900$, this set does not form a right triangle. 7. For (16, 10, 6): Longest side is 16. $$10^2 + 6^2 = 100 + 36 = 136$$ $$16^2 = 256$$ Since $136 \neq 256$, this set does not form a right triangle. 8. For (13, 5, 12): Longest side is 13. $$5^2 + 12^2 = 25 + 144 = 169$$ $$13^2 = 169$$ Since $25 + 144 = 169$, this set forms a right triangle. 9. For (10, 10, 15): Longest side is 15. $$10^2 + 10^2 = 100 + 100 = 200$$ $$15^2 = 225$$ Since $200 \neq 225$, this set does not form a right triangle. **Final answer:** The sets that can form a right triangle are: - 40, 50, 30 - 41, 9, 40 - 13, 5, 12