Question

The perimeter of a rectangular field is 628m.The length of the field exceeds its width by 6m.Find the dimensions.​

Answer 1

let the breadth of the field be X m

then length =( X + 6) m

perimeter = 2 {X + (X+6)}

= 2(2X+6) = 4X + 12

according to the question,

4X+ 12 = 628

4X = 628-12

4X = 616

X = 616/4 = 154 m

breadth= 154 m

then length = 154+6 = 160 m

Answer 2

Answer:

The perimeter is 2 times the length + 2 times the width:

2L+2W=628 L=length W=width

The width is 6m shorter than the Length:

W=L-6

Substitute W from the second equation into the first:

2L+2*(L-6)=628

2L+2L-12=628

4L=640

L=160

Now substitute this back into the second equation fo find W.

W=L-6 = 160 -6 =154

Therefore:

L=160