1. **State the problem:** Determine if each set of three side lengths can form a right triangle. 2. **Formula and rule:** Use the Pythagorean theorem: $$a^2 + b^2 = c^2$$ where $c$ is the longest side. 3. **Check each set:** - For each triple $(a,b,c)$, identify the longest side $c$. - Calculate $a^2 + b^2$ and compare to $c^2$. - If equal, it's a right triangle; if not, it isn't. 4. **Calculations:** - (6,8,10): $6^2 + 8^2 = 36 + 64 = 100$, $10^2=100$ → right triangle - (3,5,6): $3^2 + 5^2 = 9 + 25 = 34$, $6^2=36$ → not right - (12,15,20): $12^2 + 15^2 = 144 + 225 = 369$, $20^2=400$ → not right - (18,80,82): $18^2 + 80^2 = 324 + 6400 = 6724$, $82^2=6724$ → right - (4,9,11): $4^2 + 9^2 = 16 + 81 = 97$, $11^2=121$ → not right - (5,14,20): $5^2 + 14^2 = 25 + 196 = 221$, $20^2=400$ → not right - (7,24,25): $7^2 + 24^2 = 49 + 576 = 625$, $25^2=625$ → right - (5,12,13): $5^2 + 12^2 = 25 + 144 = 169$, $13^2=169$ → right - (15,20,25): $15^2 + 20^2 = 225 + 400 = 625$, $25^2=625$ → right - (30,41,49): $30^2 + 41^2 = 900 + 1681 = 2581$, $49^2=2401$ → not right - (10,24,26): $10^2 + 24^2 = 100 + 576 = 676$, $26^2=676$ → right - (16,30,34): $16^2 + 30^2 = 256 + 900 = 1156$, $34^2=1156$ → right - (2,4,5): $2^2 + 4^2 = 4 + 16 = 20$, $5^2=25$ → not right - (8,14,16): $8^2 + 14^2 = 64 + 196 = 260$, $16^2=256$ → not right - (9,12,15): $9^2 + 12^2 = 81 + 144 = 225$, $15^2=225$ → right - (10,12,13): $10^2 + 12^2 = 100 + 144 = 244$, $13^2=169$ → not right - (11,30,35): $11^2 + 30^2 = 121 + 900 = 1021$, $35^2=1225$ → not right - (20,21,29): $20^2 + 21^2 = 400 + 441 = 841$, $29^2=841$ → right - (28,45,51): $28^2 + 45^2 = 784 + 2025 = 2809$, $51^2=2601$ → not right - (36,77,85): $36^2 + 77^2 = 1296 + 5929 = 7225$, $85^2=7225$ → right 5. **Final classification:** **Right Triangles:** 6,8,10 18,80,82 7,24,25 5,12,13 15,20,25 10,24,26 16,30,34 9,12,15 20,21,29 36,77,85 **Not Right Triangles:** 3,5,6 12,15,20 4,9,11 5,14,20 30,41,49 2,4,5 8,14,16 10,12,13 11,30,35 28,45,51