The area of a rectangular football field is 2 | vedclass

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(B) Let the length of the field be $L$ and the breadth be $B$.Given that $B = L / 2$,or $L = 2B$.The area of the rectangle is $L 	imes B = 24200 \ m^

Vedclass · Accepted Answer

$246$

Vedclass · Answer

(B) Let the length of the field be $L$ and the breadth be $B$.Given that $B = L / 2$,or $L = 2B$.The area of the rectangle is $L 	imes B = 24200 \ m^2$.Substituting $L = 2B$,we get $(2B) 	imes B = 24200$,which simplifies to $2B^2 = 24200$.Dividing by $2$,we get $B^2 = 12100$,so $B = \sqrt{12100} = 110 \ m$.Then,the length $L = 2 	imes 110 = 220 \ m$.The minimum distance between opposite corners is the diagonal of the rectangle,given by $d = \sqrt{L^2 + B^2}$.$d = \sqrt{220^2 + 110^2} = \sqrt{48400 + 12100} = \sqrt{60500}$.$d = 110 	imes \sqrt{5} \approx 110 	imes 2.236 = 245.96 \ m$.Rounding to the nearest whole number,the distance is approximately $246 \ m$.

The area of a rectangular football field is $24200 \ m^2$. Its breadth is half of its length. What is the approximate minimum distance a man will cover if he wishes to go from one corner to the opposite corner? (in $m$)

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