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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?

Respuesta :

calculista

Answer:

The perimeter of rectangle is [tex]18\ cm[/tex]

Step-by-step explanation:

Let

x-----> the length of the rectangle

y----> the width of the rectangle

we 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 equation

The 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]

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