Question

Solve the following problems using two variables.
(iv) 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.

Answer 1

Let the length of rectangular yard be x m and the width of rectangular yard be y m.
According to the question, we have:
2(x + y) = 60
⇒ x + y = 30 ...(1)
and
x = y + 6
⇒ x $-$ y = 6 ...(2)
Adding (1) and (2), we get:
x + y = 30
x $-$ y = 6
----------------
2x = 36
⇒ x = 18
Substituting the value of x in equation (1), we get:
18 + y = 30
⇒ y = 12
∴ Length = 18 m and width = 12 m