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:
[tex]P=2(l+b)[/tex]Let width = b
length =l
If the length of a rectangle is 8 inches more than its width then:
[tex]l=b+8[/tex]The area of a rectangle is:
[tex]A=l\times b[/tex]If the area of the rectangle is equal to 3 inches less than 3 times the perimeter.
[tex]\begin{gathered} l\times b=3(2(l+b))-3 \\ \\ l\times b=6(l+b)-3 \end{gathered}[/tex]Put the value of "l"
[tex]\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}[/tex]Solve for "b" is:
[tex]\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}[/tex]"b" is the width of a rectangle so the value of "b" is a non-negative value.
So value of "b" is 9.
[tex]\begin{gathered} l=b+8 \\ \\ l=9+8 \\ \\ l=17 \end{gathered}[/tex]Length of rectangle is 17 and width is 9.
