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(i) The distance around the track along its inner edge.

(ii) The area of the track.

Answer :

(i) The distance around the track around along its inner edge.

= BC + EH + 2 × circumference of the semicrircle of radius 30 m

\(= \ 106 \ + \ 106 \ + \ 2 \ × \ \frac{1}{2} \ × \ 2 \pi ×30 \ \)

\( = \ 212 \ + \ 2 \ × \ \frac{22}{7} \ × \ 30 \)

\(= \ 212 \ + \ \frac{1320}{7} \ \)

\( = \ \frac{2804}{7} \)m

(ii) Area of the track

= Area of rectangle ABCD + Area of rectangle EFGH + 2 (Area of the semicircle of radius 40 m - Area of the semicircle with radius 30 m)

\(= \ (10 \ × \ 106) \ + \ (10 \ + \ 106) \ + \ 2[ \frac{1}{2} \ × \ \frac{22}{7} \ × \ (40)^2 \ - \ \frac{1}{2} \ × \ \frac{22}{7} \ × \ (30)^2] \)

\( = \ 1060 \ + \ 1060 \ + \ \frac{22}{7} (40^2 \ - \ 30^2) \ = \ 2120 \ + \ \frac{22}{7} \ × \ 70 \ × \ 10 \)

\(= \ 2120 \ + \ 2200 \ = \ 4320 \)m^{2}

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