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. **Step (i): Show $a + b > c$ given $a^2 + b^2 = c^2$** - We want to prove $a + b > c$. - Start with the inequality to prove: $$a + b > c$$ - Square both sides (valid since all sides are positive): $$ (a + b)^2 > c^2 $$ - Expand left side: $$ a^2 + 2ab + b^2 > c^2 $$ - Substitute $c^2 = a^2 + b^2$: $$ a^2 + 2ab + b^2 > a^2 + b^2 $$ - Cancel $a^2 + b^2$ on both sides: $$ \cancel{a^2} + 2ab + \cancel{b^2} > \cancel{a^2} + \cancel{b^2} $$ $$ 2ab > 0 $$ - Since $a,b > 0$, $2ab > 0$ is true. - Therefore, $a + b > c$ holds, so the sides satisfy the triangle inequality and can form a triangle. 3. **Step (ii): Prove the converse Pythagorean theorem** - Given $a,b,c > 0$ and $a^2 + b^2 = c^2$, we want to show the triangle with sides $a,b,c$ is right-angled. - By the Pythagorean theorem, if the square of the longest side equals the sum of squares of the other two sides, the triangle is right-angled. - Since $c$ is the longest side (because $c^2 = a^2 + b^2$), the angle opposite side $c$ is a right angle. - Hence, the triangle with sides $a,b,c$ is right-angled. 4. **Step (iii): Check if $(5,12,13)$ form a right triangle** - Compute $5^2 + 12^2 = 25 + 144 = 169$ - Compute $13^2 = 169$ - Since $5^2 + 12^2 = 13^2$, the sides satisfy the Pythagorean theorem. - Therefore, a triangle with sides $(5,12,13)$ is right-angled. **Final answers:** (i) $a + b > c$ is true because $2ab > 0$. (ii) The triangle with sides $a,b,c$ is right-angled if $a^2 + b^2 = c^2$. (iii) Yes, $(5,12,13)$ form a right triangle.