Length of a rectangle is 5 cm longer than the width. Four squares are constructed outside the rectangle such that each of the squares share one side with the rectangle. The total area of the constructed figure is 120 cm2. What is the perimeter of the rectangle?
Accepted Solution
A:
Answer:The perimeter of rectangle is [tex]18\ cm[/tex]Step-by-step explanation:Letx------> the length of rectangley----> the width of rectanglewe know thatThe area of the constructed figure is equal to[tex]A=xy+2x^{2} +2y^{2}[/tex][tex]A=120\ cm^{2}[/tex]so[tex]120=xy+2x^{2} +2y^{2}[/tex] -----> equation A[tex]x=y+5[/tex] -----> equation Bsubstitute equation B in equation A[tex]120=y(y+5)+2(y+5)^{2} +2y^{2}[/tex][tex]120=y^{2}+5y+2(y^{2}+10y+25)+2y^{2}[/tex][tex]120=y^{2}+5y+2y^{2}+20y+50+2y^{2}[/tex][tex]5y^{2}+25y+50-120=0[/tex][tex]5y^{2}+25y-70=0[/tex]using a graphing tool to resolve the quadratic equationthe solution is [tex]y=2\ cm[/tex]Find the value of x[tex]x=2+5=7\ cm[/tex]Find the perimeter of rectangleThe perimeter of rectangle is equal to[tex]P=2(x+y)[/tex][tex]P=2(7+2)=18\ cm[/tex]