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The perimeter of the rectangular playing field is 290 yards. The length of the field is 8 yards less than double the width. How would you find the dimensions of the playing​ field?

Respuesta :

Let width be x

Length=2x-8

We know

[tex]\boxed{\sf Perimeter=2(L+B)}[/tex]

[tex]\\ \sf\longmapsto 2(x+2x-8)=290[/tex]

[tex]\\ \sf\longmapsto 2(3x-8)=290[/tex]

[tex]\\ \sf\longmapsto 3x-8=\dfrac{290}{2}[/tex]

[tex]\\ \sf\longmapsto 3x-8=145[/tex]

[tex]\\ \sf\longmapsto 3x=145+8[/tex]

[tex]\\ \sf\longmapsto 3x=153[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{153}{3}[/tex]

[tex]\\ \sf\longmapsto x=51[/tex]

Length=2x-8=2(51)-8=102-8=96

Width=x=51

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