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# 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

- PinkgreenLv 71 month ago
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.

- KrishnamurthyLv 72 months ago
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.

- Blue JackLv 62 months ago
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

- PuzzlingLv 72 months ago
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.

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- Daniel HLv 52 months ago
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

- 2 months ago
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.