the perimeter of a rectangle is 30 ft. The width is 5 ft more than the length. What's the width and length?

Answer 1

The perimeter of a rectangle is 30 ft. The width is 5 ft more than the length. What's the width and length?

=>

width = w

length = l = w+5

2(l+w) =p

2(w+5+w) = 30

2(2w+5)=30

2w +5 =15

2w =10

w =5

width = w= 5 ft

length = l = w+5 =5+5 =10 ft

Answer 2

We know that the perimeter of a rectangle is the summation of all of the sides. There are four sides of a rectangle, but two sets of the same size. So we can write the perimeter (P) as the summation of the lengths (L) and the widths (W), or:

P = L + L + W + W = 2L + 2W.

We were given two pieces of information: 1. that the perimeter (P) of the rectangle is 30 ft, and 2. that the width (W) is 5 ft more than the length (L). We can express this information with these two equations:

P = 2L + 2W = 30ft

W = L + 5ft

Now, we have two unknowns. We don't know the length (L) or the width (W) of the rectangle, but we have two linearly independent equations, which is the same number of unknowns that we have, so we can solve for the width (W) and the length (L) of the rectangle.

W = L + 5ft

2L + 2W = 30ft

Using the W = L + 5ft equation, we can plug that into the perimeter equation and get:

2L + 2(L + 5ft) = 30ft

Now divide both sides of the equation by 2:

L + (L + 5ft) = 15ft

Grouping common terms:

2L + 5ft = 15ft

Subtracting 5ft from both sides of the equation:

2L + 5ft - 5ft = 15ft - 5ft

2L = 10ft

And dividing both sides by 2 gives:

L = 5ft.

Now, using the new found value for L, and using the relation W = L + 5ft, we can solve for the width (W):

W = (5ft) + 5ft

W = 10ft.

So we have found the length (L) and the width (W) of the rectangle, and they are:

L = 5ft

W = 10ft