contestada

A rectangle has a perimeter of 34 cm in an area of 52 cm squared it's length is 5 more than twice its width write and solve the system of equations to find the dimensions of the rectangle

Respuesta :

L+W=17 W=4 is the answer.

Let [tex] l [/tex] be the length and [tex] w [/tex] be the width of the rectangle.

Given perimeter of the rectangle

[tex] 2(l+w)=34 \; l+w=17[/tex]

Area of the rectangle is

[tex] lw=52 [/tex]

Also length is 5 more than twice its width,

[tex] l=2w+5 [/tex]

From the above equations

[tex] (l-w)^2=(l+w)^2-4lw=17^2-4*52=81\\
l-w=9,\; since \; l>w [/tex]

Now [tex] 2l=9+17=26\;,l=13,w=4 [/tex]

The dimensions of the rectangle are length [tex] l=13 [/tex], and width [tex] w=4 [/tex].


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