The perimeter of rectrangular field is 172 meters. Double the width is 4 meters more than the length. Find the area of the field.

I would like to know what is the equation to use to solve the problem.

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The perimeter of rectangular field is 172 meters.

Let the width of the field be W = X. Twice the width of the field is 4 more than the length. The length is equal to L = 2*X - 4.

The perimeter of the rectangular field is 2*(W + L)

=> 2*(X + 2*X - 4)

=> 6x - 8

6X - 8 = 172

=> X = 30

**The area of the field is W*L = 30*56 = 1680 square meters.**

perimeter=2(l+b)

perimeter=172 m

area=lb

l=2w -4

perimeter=2({2w-4}+w)

=4w-8+2w

=6w-8

172=6w-8

6w=172+8

=180

w=180/6

=30m

l=(2x30)-4

=60-4

=56m

area=56x30

=1680 sq. m (ans)

Here w is width and l is length

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