Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

# Math Question?

The length of a rectangular pool is to be 2 times its width, and a sidewalk 6.5 feet wide will surround the pool. If a total area of 1809 ft2 has been set aside for construction, what are the dimensions of the pool?

### 6 Answers

Relevance
• Let w ft be the width of the pool, then 2w

will be the length. 1809 ft^2 is the total area

of the pool & the sidewalk, you should first

write down the true length & width of the

rectangle for this area which are

length=2w+2(6.5)

width=w+2(6.5)

& then write down an equation & solve for w

& hence get the dimensions of the pool.

Ans. w=20.5 ft; 2w=41 ft.

• The length of a rectangular pool is to be 2 times its width,

and a sidewalk 6.5 feet wide will surround the pool.

If a total area of 1809 ft^2 has been set aside for construction,

what are the dimensions of the pool?

(2w + 13)*(w + 13) = 1809

w = 41/2, l = 41

The length of the pool is 41 feet and its width is 20 1/2 feet.

• A = 1809 ft²

s = 6.5 ft

l = pool length

w = pool width

l = 2w

A = (2w + 2s)(w + 2s)

1809 = (2w + 2(6.5))(w + 2(6.5)

1809 = 2w² + 13w + 26w + 13²

2w² + 39w - 1640 = 0

If ax² + bx + c = 0 then x = [-b ± √(b² - 4ac)] / 2a

w = [-39 ± √(39² - 4(2)(-1640))] / 2(2)

w = -39/4 ± 121/4

w = 20.5, -40

Throw out the negative value which makes no sense.

w = 20.5 ft

l = 2w

l = 2(20.5)

l = 41 ft

• Let w be the width.

Let 2w be the length.

The total width of the area is w + 2(6.5) --> w + 13

The total length of the area is 2w + 2(6.5) --> 2w + 13

The total  area is 1809 ft²:

(w + 13)(2w + 13) = 1809

Expand the left side:

2w² + 13w + 26w + 169 = 1809

2w² + 39w - 1640 = 0

Factor that:

(2w - 41)(w + 40) = 0

That leads to two potential solutions:

2w - 41 = 0

2w = 41

w = 20.5

w + 40 = 0

w = -40

We can ignore the negative result.

Calculate the length (double the width):

2w = 2(20.5) = 41

Answer:

The pool is 20.5 feet wide and 41 feet long.

• How do you think about the answers? You can sign in to vote the answer.
• pool

width = w

length = 2w

entier = pool plus border sidwalk

width = w + 2(6.5ft) = w + 13ft

length = 2w + 2(6.5ft) = 2w + 13ft

Area = 1809ft^2

1809ft^2 = (w + 13ft)(2w + 13ft)

1809ft^2 = 2w^2 + 39ftw + 169ft^2

2w^2 + 39ftw - 1640ft

w = 20.5ft and/or -40ft

ignore negative width

pool measurements

w = 20.5ft

l = 41ft

verify

(41 + 13)(20.5 + 13) = 1809ft

• Let's say the dimensions of the pool are l and w. imagine a "frame" of 6.5 ft added around the pool. the dimensions of the new rectangle are l + 6.5 + 6.5 (6.5 on each side) and w + 6.5 + 6.5 or in short l + 13 and w + 13. We know the total surface is 1809, so (l + 13)(w + 13) = 1809.

We also know that the length is twice the width, so we could say l = 2w. This means (2w + 13)(w + 13) = 1809. or 2w² + 39w + 169 = 1809 or 2w² + 39w - 1640 = 0.

Solve this equation and you'll find your dimensions.

Still have questions? Get your answers by asking now.