The length of a rectangle is 9 ft longer than twice the width. if the perimeter is 108 ft, find the length and the width of the rectangle

Let the width of rectangle be x ft So length of rectangle be (2x+ 9 ) ft Given Perimeter of rectangle =2 (l+b) Or 108= 2 (2x+9+x) Or 108÷2=3x+9 Or 54=3x+9 Or 54-9=3x Or 45=3x Or x= 15 Required length = (2x+9)=(2×15+9)=39 ft And width =x= 15 ft

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