Q. 83.9( 365 Votes )

# Fig. 12.26 depict

(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 πr2
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 m2

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