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.