1. **Problem statement:** We have an oval track ABCD with given lengths |AB| = 20 cm and |BC| = 14 cm. We need to calculate: (i) The perimeter of the track. (ii) The area of the track. Take $\pi = \frac{22}{7}$. 2. **Understanding the shape:** The track is an oval, which can be approximated as an ellipse with major axis length $2a = 20$ cm and minor axis length $2b = 14$ cm. So, $a = \frac{20}{2} = 10$ cm and $b = \frac{14}{2} = 7$ cm. 3. **Formulas:** - Perimeter (circumference) of an ellipse does not have a simple exact formula, but Ramanujan's approximation is very accurate: $$ P \approx \pi \left[ 3(a+b) - \sqrt{(3a + b)(a + 3b)} \right] $$ - Area of an ellipse: $$ A = \pi a b $$ 4. **Calculate the perimeter:** Calculate each term: $3(a+b) = 3(10 + 7) = 3 \times 17 = 51$ $3a + b = 3 \times 10 + 7 = 30 + 7 = 37$ $a + 3b = 10 + 3 \times 7 = 10 + 21 = 31$ Calculate the square root term: $$ \sqrt{37 \times 31} = \sqrt{1147} $$ Approximate $\sqrt{1147} \approx 33.87$ (rounded to two decimals). Now substitute into the perimeter formula: $$ P \approx \pi (51 - 33.87) = \pi \times 17.13 $$ Using $\pi = \frac{22}{7} \approx 3.142857$: $$ P \approx 3.142857 \times 17.13 \approx 53.82 \text{ cm} $$ 5. **Calculate the area:** $$ A = \pi a b = \frac{22}{7} \times 10 \times 7 $$ Simplify: $$ A = \frac{22}{7} \times \cancel{10} \times \cancel{7} = 22 \times 10 = 220 \text{ cm}^2 $$ 6. **Final answers:** - Perimeter $\approx 53.82$ cm - Area $= 220$ cm$^2$