The figure depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is $60 \, m$ and they are each $106 \, m$ long. If the track is $10 \, m$ wide,find:
$(i)$ the distance around the track along its inner edge
$(ii)$ the area of the track. $\left[ \text{Use } \pi = \frac{22}{7} \right]$

(N/A) Given:
Inner length of the straight parts $= 106 \, m$
Inner width (distance between parallel lines) $= 60 \, m$
Inner radius $(r) = \frac{60}{2} = 30 \, m$
Width of the track $= 10 \, m$
Outer radius $(R) = 30 + 10 = 40 \, m$
$(i)$ Distance around the track along its inner edge $= AB + \text{arc } BEC + CD + \text{arc } DFA$
$= 106 + (\pi r) + 106 + (\pi r)$
$= 212 + 2 \pi r$
$= 212 + 2 \times \frac{22}{7} \times 30$
$= 212 + \frac{1320}{7} = \frac{1484 + 1320}{7} = \frac{2804}{7} \, m \approx 400.57 \, m$
$(ii)$ Area of the track $= 2 \times (\text{Area of rectangle } 106 \times 10) + 2 \times (\text{Area of semi-circular ring})$
$= 2 \times (106 \times 10) + 2 \times \left[ \frac{1}{2} \pi (R^2 - r^2) \right]$
$= 2120 + \pi (40^2 - 30^2)$
$= 2120 + \frac{22}{7} (1600 - 900)$
$= 2120 + \frac{22}{7} \times 700$
$= 2120 + 2200 = 4320 \, m^2$