4. A house has rectangular yard in front of it for children to play. The length of that rectangle exceeds its width by 6 m and its perimeter is 60 m, find the measurement of the yard.



Solution. Let the length of the rectangular yard be x m and its width be y m.

As per the first given condition,
x = y + 6
∴  x – y = 6 ........eq. no. (1)

Perimeter of the rectangular yard = 2 (l + b) = 2 (x + y) m

As per the second given condition,

2 (x + y) = 60

∴ x + y = 30 ......eq, no. (2)

Adding (1) and (2), we get,

x – y = 6
x + y = 30
2x     = 36
∴ x = 36/2
∴ x = 18
Substituting x = 18 in equation no. (2)
∴ x + y = 30
∴ 18 + y = 30
∴ y = 30 – 18
∴ y = 12


The length and width of the yard are 18 m and 12 m respectively.

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