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(B) Length of the outer rectangle $= 60 \ m$.Breadth of the outer rectangle $= 40 \ m$.Area of the outer rectangle $= 60 \ m 	imes 40 \ m = 2400 \ m^

Vedclass · Accepted Answer

$196$

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(B) Length of the outer rectangle $= 60 \ m$.Breadth of the outer rectangle $= 40 \ m$.Area of the outer rectangle $= 60 \ m 	imes 40 \ m = 2400 \ m^2$.Width of the path $= 1 \ m$.Since the path is inside the lawn,the length of the inner rectangle $= 60 \ m - (1 \ m + 1 \ m) = 58 \ m$.The breadth of the inner rectangle $= 40 \ m - (1 \ m + 1 \ m) = 38 \ m$.Area of the inner rectangle $= 58 \ m 	imes 38 \ m = 2204 \ m^2$.Area of the path $= 	ext{Area of outer rectangle} - 	ext{Area of inner rectangle}$.Area of the path $= 2400 \ m^2 - 2204 \ m^2 = 196 \ m^2$.

$A$ lawn is in the shape of a rectangle of length $60 \ m$ and width $40 \ m$. Inside the lawn,there is a footpath of uniform width $1 \ m$ bordering the lawn. The area of the path is ....... $m^2$.

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