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12.3 ka 8 no

Posted by Alisha. Afreen 4 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Alisha. Afreen 4 years, 9 months ago

Thanks ?

0Thank You

Gaurav Seth 4 years, 9 months ago

Width of the track = 10 m

Distance between two parallel lines = 60 m

Length of parallel tracks = 106 m

DE = CF = 60 m

Radius of inner semicircle, r = OD = O’C

= 60/2 m = 30 m

Radius of outer semicircle, R = OA = O’B

= 30+10 m = 40 m

Also, AB = CD = EF = GH = 106 m

Distance around the track along its inner edge = CD+EF+2×(Circumference of inner semicircle)

= 106+106+(2×πr) m = 212+(2×22/7×30) m

= 212+1320/7 m = 2804/7 m

Area of the track = Area of ABCD + Area EFGH + 2 × (area of outer semicircle) – 2 × (area of inner semicircle)

= (AB×CD)+(EF×GH)+2×(πr²/2) -2×(πR²/2) m²

= (106×10)+(106×10)+2×π/2(r²-R²) m²

= 2120+22/7×70×10 m²

= 4320 m²

1Thank You

ANSWER

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