1. The problem asks to find the range of possible lengths for the third side $x$ of a triangle given two sides, using the triangle inequality rule: $$|a-b| < x < a+b$$ where $a$ and $b$ are the given side lengths. 2. For each pair: - No. 6: sides 6 and 9 $$|6-9| < x < 6+9$$ $$3 < x < 15$$ - No. 7: sides 10 and 4 $$|10-4| < x < 10+4$$ $$6 < x < 14$$ - No. 8: sides 7 and 11 $$|7-11| < x < 7+11$$ $$4 < x < 18$$ - No. 9: sides 5 and 13 $$|5-13| < x < 5+13$$ $$8 < x < 18$$ - No. 10: sides 8 and 15 $$|8-15| < x < 8+15$$ $$7 < x < 23$$ 3. Challenge question: Given sides $x$, $x+2$, and 10 cm, find integer values of $x$ that make a triangle. Apply the triangle inequality for each pair: - Between $x$ and $x+2$: $$|(x) - (x+2)| < 10 < x + (x+2)$$ $$2 < 10 < 2x + 2$$ - From $$2 < 10$$ no restriction on $x$ here. - From $$10 < 2x + 2$$ solve: $$10 - 2 < 2x$$ $$8 < 2x$$ $$4 < x$$ - Between $x$ and 10: $$|x - 10| < x + 2 < x + 10$$ The inequalities: $$|x - 10| < x + 2$$ For $x<10$: $$10 - x < x + 2 ightarrow 10 - x < x + 2 ightarrow 8 < 2x ightarrow 4 < x$$ For $x o 10^+$ is always valid. - Between $x+2$ and 10: $$|(x+2) - 10| < x < (x+2) + 10$$ $$|x - 8| < x < x + 12$$ Focus on $$|x - 8| < x$$: For $x < 8$: $$8 - x < x ightarrow 8 < 2x ightarrow 4 < x$$ For $x o 8^+$ true. 4. Combining: $$4 < x < 8$$ to satisfy all inequalities. Integer values of $x$ are 5, 6, 7. Final answers: - No. 6: $3 < x < 15$ - No. 7: $6 < x < 14$ - No. 8: $4 < x < 18$ - No. 9: $8 < x < 18$ - No. 10: $7 < x < 23$ - Challenge: integer values for $x$ are $5, 6,$ and $7$