Answer :

**(i)** Distance around the track along its inner edge = AB + semicircle BEC + CD + semicircle DFA

= 106 + ( × 2πr )+ 106 +( × 2πr) [Perimeter of semicircle is half of circle and perimeter of circle is 2πr)

= 212 +( × 2 × × 30) +( × 2 × × 30)

= 212 + (2 × ×30)

=

= m

**(ii)** Area of the track = (Area of GHIJ - Area of ABCD) + (Area of semi-circle HKI - Area of semi-circle BEC) + (Area of semi-circle GLJ - Area of semi-circle AFD)

Area of semicircle is half of circle and area of circle is πr^{2}

Also area of rectangle = length x breadth

= (106 × 80 – 106 × 60 )+( × × (40)^{2} - × × (30)^{2}) +( × × (40)^{2} - × × (30)^{2} )

=[ 106 (80 – 60) ]+[ × (40)^{2} - × (30)^{2}]

= [106 (20)] +[ (40)^{2} – (30)^{2}]

= 2120 +[ (40 – 30) (40 + 30)]

= 2120 + (22/7) (10) (70)

= 2120 + 2200

= 4320 m^{2}

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