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