Question

The length of a rectangle is 8 inches more than its width. The area of the rectangle is equal to 3 inches less than 3 times the perimeter. Find the length and width of the rectangle.

Answer 1

Given:

The length of a rectangle is 8 inches more than its width.

The area of the rectangle is equal to 3 inches less than 3 times the perimeter.

Find-: Length and width of the rectangle.

Sol-:

The perimeter of a rectangle is:

`P=2(l+b)`

Let width = b

length =l

If the length of a rectangle is 8 inches more than its width then:

`l=b+8`

The area of a rectangle is:

`A=l* b`

If the area of the rectangle is equal to 3 inches less than 3 times the perimeter.

`\begin{gathered} l* b=3(2(l+b))-3 \\ \\ l* b=6(l+b)-3 \end{gathered}`

Put the value of "l"

`\begin{gathered} b(b+8)=6(b+8+b)-3 \\ \\ b^2+8b=6(2b+8)-3 \\ \\ b^2+8b=12b+48-3 \\ \\ b^2+8b-12b=45 \\ \\ b^2-4b-45=0 \end{gathered}`

Solve for "b" is:

`\begin{gathered} b^2-4b-45=0 \\ \\ b^2-9b+5b-45=0 \\ \\ b(b-9)+5(b-9)=0 \\ \\ (b-9)(b+5)=0 \\ \\ b=9;b=-5 \end{gathered}`

"b" is the width of a rectangle so the value of "b" is a non-negative value.

So value of "b" is 9.

`\begin{gathered} l=b+8 \\ \\ l=9+8 \\ \\ l=17 \end{gathered}`

Length of rectangle is 17 and width is 9.