1. **Stating the problem:** We have a pattern of rhombuses arranged in rows with the number of rhombuses in each pattern given as 1, 3, 5, 7 for patterns 1, 2, 3, and 4 respectively. 2. **Observing the pattern:** The number of rhombuses in each pattern forms the sequence 1, 3, 5, 7, ... which are consecutive odd numbers. 3. **Finding the number of rhombuses in pattern 5 and 6:** - Pattern 5 will have the 5th odd number: $2 \times 5 - 1 = 9$ - Pattern 6 will have the 6th odd number: $2 \times 6 - 1 = 11$ 4. **Formula for the number of rhombuses in the $n^{th}$ pattern:** The $n^{th}$ odd number is given by: $$T_n = 2n - 1$$ 5. **Number of rhombuses in the 10th pattern:** Using the formula: $$T_{10} = 2 \times 10 - 1 = 19$$ 6. **Finding which figure has 57 rhombuses:** Set $T_n = 57$: $$2n - 1 = 57$$ $$2n = 58$$ $$n = 29$$ So, the 29th figure has 57 rhombuses. 7. **Which figure has 13 rhombuses in the bottom row:** Set $T_n = 13$: $$2n - 1 = 13$$ $$2n = 14$$ $$n = 7$$ So, the 7th figure has 13 rhombuses in the bottom row. **Final answers:** - Pattern 5 has 9 rhombuses. - Pattern 6 has 11 rhombuses. - Formula: $T_n = 2n - 1$ - Pattern 10 has 19 rhombuses. - Pattern 29 has 57 rhombuses. - Pattern 7 has 13 rhombuses in the bottom row.