Question

()) A rectangular piece of paper has a perimeter of 22 inches and an area of 28 square

inches. What are the dimensions of the paper?

Answer 1

Given:

Perimeter of a rectangular paper = 22 inches.

Area of the rectangular paper = 28 square inches.

To find:

The dimensions of the rectangular paper.

Solution:

Let l be the length and w be the width of the rectangular paper.

Perimeter of a rectangle is:

`P=2(l+w)`

Perimeter of a rectangular paper is 22 inches.

`2(l+w)=22`

`l+w=(22)/(2)`

`l=11-w` ...(i)

Area of a rectangle is:

`A=lw`

Area of the rectangular paper is 28 square inches.

`28=lw`

Using (i), we get

`28=(11-w)w`

`28=11w-w^2`

`w^2-11w+28=0`

Splitting the middle term, we get

`w^2-7w-4w+28=0`

`w(w-7)-4(w-7)=0`

`(w-7)(w-4)=0`

Using zero product property, we get

`(w-7)=0\text{ and }(w-4)=0`

`w=7\text{ and }w=4`

If
`w=7`, then by using (i)

`l=11-7`

`l=4`

If
`w=4`, then by using (i)

`l=11-4`

`l=7`

Therefore, the dimensions of the paper are either
`7* 4` or
`4* 7`.