Length of a rectangle is 5 cm longer than the width. Four squares are constructed outside the rectangle such that each of the squares shares one side with the rectangle. The total area of the constructed figure is 120 cm2. What is the perimeter of the rectangle?

Answer:The perimeter of rectangle is [tex]18\ cm[/tex]Step-by-step explanation:Letx-----> the length of the rectangley----> the width of the rectanglewe know that[tex]x=y+5[/tex] ----> equation A[tex]120=xy+2x^{2}+2y^{2}[/tex] ---> equation B (area of the constructed figure)substitute the equation A in equation B[tex]120=(y+5)y+2(y+5)^{2}+2y^{2}[/tex][tex]120=(y+5)y+2(y+5)^{2}+2y^{2}\\ 120=y^{2}+5y+2(y^{2}+10y+25)+2y^{2}\\ 120=y^{2}+5y+2y^{2}+20y+50+2y^{2}\\120=5y^{2}+25y+50\\5y^{2}+25y-70=0[/tex]using a graphing calculator -----> solve the quadratic equationThe solution is[tex]y=2\ cm[/tex]Find the value of x[tex]x=y+5 ----> x=2+5=7\ cm[/tex]Find the perimeter of rectangle[tex]P=2(x+y)=2(7+2)=18\ cm[/tex]