Question

the length of a rectangle is triple the width. It's perimeter is more than 40m. Find the smallest possible values for its length and width.

Answer 1

Given:- the length of a rectangle is triple the width. it's perimeter is more than 40m.

Now, from the given condition we have ,
`l=3b` where l is the length of the rectangle and b is the width of the rectangle.

Perimeter(P) of a rectangle is given by,
`P=2(l+b)`

Since perimeter is more than 40 m we have,

`2(l+b)>40`

`2(3b+b)>40`

`2\cdot 4b>40`

8b>40

⇒ b>5 m

since,
`l=3b` then
`l>3\cdot 5`

⇒
`l >15 m`

if we take the integers, the smallest possible values for its length and width ,
`l=16` m and b=6 m.