1. **Problem statement:** (i) Show that if $a, b, c > 0$ and $a^2 + b^2 = c^2$, then $a + b > c$ so that $a, b, c$ can form a triangle. (ii) Prove the converse Pythagorean theorem: If $a, b, c > 0$ and $a^2 + b^2 = c^2$, then the triangle with sides $a, b, c$ is right-angled. (iii) Determine if a triangle with sides $(5, 12, 13)$ is right-angled. 2. **Key formula and rule:** - Triangle inequality: For any triangle with sides $a, b, c$, the sum of any two sides must be greater than the third, e.g., $a + b > c$. - Pythagorean theorem: In a right triangle, $a^2 + b^2 = c^2$ where $c$ is the hypotenuse. --- ### Part (i): Show $a + b > c$ 3. Given $a, b, c > 0$ and $a^2 + b^2 = c^2$. 4. Start with the triangle inequality to prove: $$a + b > c$$ 5. Square both sides (valid since all positive): $$ (a + b)^2 > c^2 $$ 6. Expand left side: $$ a^2 + 2ab + b^2 > c^2 $$ 7. Substitute $c^2 = a^2 + b^2$: $$ a^2 + 2ab + b^2 > a^2 + b^2 $$ 8. Cancel $a^2 + b^2$ on both sides: $$ \cancel{a^2} + 2ab + \cancel{b^2} > \cancel{a^2} + \cancel{b^2} $$ 9. Simplify: $$ 2ab > 0 $$ 10. Since $a, b > 0$, $2ab > 0$ is true. 11. Therefore, $a + b > c$ holds, so $a, b, c$ satisfy the triangle inequality and can form a triangle. --- ### Part (ii): Converse Pythagorean theorem 12. Given $a, b, c > 0$ and $a^2 + b^2 = c^2$, prove the triangle with sides $a, b, c$ is right-angled. 13. By the triangle inequality from part (i), $a, b, c$ form a triangle. 14. The converse Pythagorean theorem states: If $a^2 + b^2 = c^2$, then the angle opposite side $c$ is a right angle. 15. Using the Law of Cosines: $$ c^2 = a^2 + b^2 - 2ab \cos(\theta) $$ where $\theta$ is the angle opposite side $c$. 16. Substitute $c^2 = a^2 + b^2$: $$ a^2 + b^2 = a^2 + b^2 - 2ab \cos(\theta) $$ 17. Cancel $a^2 + b^2$ on both sides: $$ 0 = - 2ab \cos(\theta) $$ 18. Divide both sides by $-2ab$ (nonzero since $a,b>0$): $$ \cancel{\frac{0}{-2ab}} = \cancel{\frac{-2ab \cos(\theta)}{-2ab}} $$ 19. Simplify: $$ 0 = \cos(\theta) $$ 20. Therefore: $$ \theta = 90^\circ $$ 21. Hence, the triangle with sides $a, b, c$ is right-angled at the angle opposite side $c$. --- ### Part (iii): Check if $(5, 12, 13)$ form a right triangle 22. Compute $5^2 + 12^2$: $$ 25 + 144 = 169 $$ 23. Compute $13^2$: $$ 169 $$ 24. Since $5^2 + 12^2 = 13^2$, by the Pythagorean theorem, the triangle with sides $(5, 12, 13)$ is right-angled. 25. Also, $5 + 12 = 17 > 13$, so the triangle inequality holds. **Final answers:** - (i) $a + b > c$ is true, so $a, b, c$ form a triangle. - (ii) If $a^2 + b^2 = c^2$, the triangle is right-angled. - (iii) The triangle with sides $(5, 12, 13)$ is right-angled.