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(D) Let the length of the rectangle be $l$ and the width be $w$.Diagonal $d = \sqrt{l^2 + w^2}$.Distance covered by $A$ diagonally $= \frac{52 	ext{

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(D) Let the length of the rectangle be $l$ and the width be $w$.Diagonal $d = \sqrt{l^2 + w^2}$.Distance covered by $A$ diagonally $= \frac{52 	ext{ m}}{60 	ext{ s}} 	imes 15 	ext{ s} = 13 	ext{ m}$.So,$l^2 + w^2 = 13^2 = 169$.Distance covered by $B$ along the sides $= l + w = \frac{68 	ext{ m}}{60 	ext{ s}} 	imes 15 	ext{ s} = 17 	ext{ m}$.We know that $(l + w)^2 = l^2 + w^2 + 2lw$.Substituting the values: $17^2 = 169 + 2lw$.$289 = 169 + 2lw$.$2lw = 289 - 169 = 120$.Area of the field $= l 	imes w = \frac{120}{2} = 60 	ext{ m}^2$.

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