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The rectangle shown has a perimeter of 34 cm and the area given its length is 5 more than twice the width write and solve a system of equations to find the dimensions of the rectangle a=52cm^2

Respuesta :

juniorherrea45
2(2w+5) + 2W =34
4w+10+2w=34
6W+10=34
-10 -10
6w=24
Width =4
Length =13

Answer:

The width and the length of rectangle are 4 cm and 13 cm respectively .

Step-by-step explanation:

Let the width be x

We are given that its length is 5 more than twice the width

So, length= 2x+5

Perimeter of rectangle = [tex]2(L+W)[/tex]

                                    = [tex]2(x+2x+5)[/tex]  

We are given that perimeter is 34 cm

So, [tex]2(x+2x+5)= 34[/tex]  

[tex]6x+10= 34[/tex]  

[tex]6x= 24[/tex]  

x=4

So, width = 4 cm

Length = 2x+5=2(4)+5=13 cm

Area of rectangle = [tex]Length \times width = x(2x+5)[/tex]

We are given that area is 52sq.cm

So, [tex]x(2x+5) =52[/tex]

[tex]2x^2+5x=52[/tex]

[tex](x-4)(2x+13)=0[/tex]

[tex]x=4,\frac{-13}{2}[/tex]

Since width cannot be negative

So,width = 4 cm

Length = 2x+5=2(4)+5=13 cm

Hence The width and the length of rectangle are 4 cm and 13 cm respectively .

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