1. **Problem statement:** A rectangular garden has a diagonal of 20 m and length of 16 m. A circular pond is constructed inside it. We need to find: (a) The perimeter of the garden. (b) The area of the pond if its diameter is $d$ meters. (c) The area occupied by the pond if the breadth of the garden is 5 m more than the diameter of the pond. 2. **Formulas and rules:** - For a rectangle with length $l$ and breadth $b$, the diagonal $d$ satisfies $$d^2 = l^2 + b^2$$ - The perimeter $P$ of a rectangle is $$P = 2(l + b)$$ - The area $A$ of a circle with diameter $d$ is $$A = \pi \frac{d^2}{4}$$ 3. **Step (a): Find the breadth $b$ and perimeter $P$ of the garden.** Given diagonal $d = 20$ m and length $l = 16$ m. Using Pythagoras theorem: $$20^2 = 16^2 + b^2$$ $$400 = 256 + b^2$$ $$b^2 = 400 - 256 = 144$$ $$b = \sqrt{144} = 12 \text{ m}$$ Now, perimeter: $$P = 2(l + b) = 2(16 + 12) = 2 \times 28 = 56 \text{ m}$$ 4. **Step (b): Area of the pond with diameter $d$.** The area of the pond is: $$A = \pi \frac{d^2}{4}$$ 5. **Step (c): Find the area of the pond if breadth $b$ is 5 m more than diameter $d$.** Given: $$b = d + 5$$ From step (a), we know $b = 12$ m, so: $$12 = d + 5$$ $$d = 12 - 5 = 7 \text{ m}$$ Now, area of pond: $$A = \pi \frac{7^2}{4} = \pi \frac{49}{4} = \frac{49\pi}{4} \approx 38.48 \text{ m}^2$$ **Final answers:** (a) Perimeter of the garden = $56$ m (b) Area of the pond = $\pi \frac{d^2}{4}$ (c) Area of the pond when $d=7$ m = approximately $38.48$ m$^2$