1. **Problem 1:** We have two rectangles. - Left rectangle: 3 columns and 4 rows. - Right rectangle: 10 columns and 4 rows. Each small square has an area of 1 cm². 2. **Formulas:** - Perimeter of a rectangle: $$P = 2(l + w)$$ where $l$ is length and $w$ is width. - Area of a rectangle: $$A = l \times w$$ 3. **Calculations for Problem 1:** - Left rectangle: - Length $l = 4$ cm (rows), Width $w = 3$ cm (columns). - Perimeter: $$P = 2(4 + 3) = 2 \times 7 = 14\text{ cm}$$ - Area: $$A = 4 \times 3 = 12\text{ cm}^2$$ - Right rectangle: - Length $l = 4$ cm, Width $w = 10$ cm. - Perimeter: $$P = 2(4 + 10) = 2 \times 14 = 28\text{ cm}$$ - Area: $$A = 4 \times 10 = 40\text{ cm}^2$$ 4. **Draw another rectangle with the same perimeter as the left rectangle (14 cm):** - Let’s find dimensions $l$ and $w$ such that $2(l + w) = 14$. - Simplify: $$l + w = 7$$ - Possible pairs: (1,6), (2,5), (3,4), etc. - For example, $l=5$, $w=2$. - Area: $$A = 5 \times 2 = 10\text{ cm}^2$$ 5. **Draw another rectangle with the same perimeter as the right rectangle (28 cm):** - $$l + w = 14$$ - Possible pairs: (6,8), (7,7), (5,9), etc. - For example, $l=7$, $w=7$. - Area: $$A = 7 \times 7 = 49\text{ cm}^2$$ 6. **Problem 2:** - Left rectangle: 2 columns and 4 rows. - Right rectangle: 8 columns and 3 rows. 7. **Calculations for Problem 2:** - Left rectangle: - Length $l=4$, Width $w=2$. - Perimeter: $$P = 2(4 + 2) = 2 \times 6 = 12\text{ cm}$$ - Area: $$A = 4 \times 2 = 8\text{ cm}^2$$ - Right rectangle: - Length $l=3$, Width $w=8$. - Perimeter: $$P = 2(3 + 8) = 2 \times 11 = 22\text{ cm}$$ - Area: $$A = 3 \times 8 = 24\text{ cm}^2$$ 8. **Draw another rectangle with the same perimeter as the left rectangle (12 cm):** - $$l + w = 6$$ - Possible pairs: (1,5), (2,4), (3,3), etc. - For example, $l=5$, $w=1$. - Area: $$A = 5 \times 1 = 5\text{ cm}^2$$ 9. **Draw another rectangle with the same perimeter as the right rectangle (22 cm):** - $$l + w = 11$$ - Possible pairs: (5,6), (4,7), (3,8), etc. - For example, $l=6$, $w=5$. - Area: $$A = 6 \times 5 = 30\text{ cm}^2$$ 10. **Garden problem:** - Area $A = 5$ m². - We want to find the perimeter $P$. - Let length be $l$ and width be $w$. - We know: $$l \times w = 5$$ - Perimeter: $$P = 2(l + w)$$ - Without specific dimensions, perimeter is not unique. - For example, if $l=5$, $w=1$, then $$P = 2(5 + 1) = 12\text{ m}$$ - If $l=\sqrt{5}$, $w=\sqrt{5}$ (square), then $$P = 2(\sqrt{5} + \sqrt{5}) = 4\sqrt{5} \approx 8.94\text{ m}$$ - So, perimeter depends on the shape, but area fixed at 5 m². **Final answers:** 1. - Left rectangle: Perimeter = 14 cm, Area = 12 cm² - Right rectangle: Perimeter = 28 cm, Area = 40 cm² - New rectangle with perimeter 14 cm: Area = 10 cm² - New rectangle with perimeter 28 cm: Area = 49 cm² 2. - Left rectangle: Perimeter = 12 cm, Area = 8 cm² - Right rectangle: Perimeter = 22 cm, Area = 24 cm² - New rectangle with perimeter 12 cm: Area = 5 cm² - New rectangle with perimeter 22 cm: Area = 30 cm² 3. - Garden area = 5 m² - Perimeter depends on dimensions, e.g., 12 m or approximately 8.94 m depending on shape.