Question

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?

Answer 1

Let width be x

Length=2x-8

We know

`\boxed{\sf Perimeter=2(L+B)}`

`\\ \sf\longmapsto 2(x+2x-8)=290`

`\\ \sf\longmapsto 2(3x-8)=290`

`\\ \sf\longmapsto 3x-8=(290)/(2)`

`\\ \sf\longmapsto 3x-8=145`

`\\ \sf\longmapsto 3x=145+8`

`\\ \sf\longmapsto 3x=153`

`\\ \sf\longmapsto x=(153)/(3)`

`\\ \sf\longmapsto x=51`

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

Width=x=51