# Dimension of the poolA rectangular pool is surrounded by a walk 4 feet wide. The pool is 6 feet longer that it is wide. If the area of the pool and the walk is 272 square feet more than the area of...

A rectangular pool is surrounded by a walk 4 feet wide. The pool is 6 feet longer that it is wide. If the area of the pool and the walk is 272 square feet more than the area of the pool. What are the dimensions of the pool?

I really dont get the right answer :( , please help me to understand this. I need the solution

*print*Print*list*Cite

### 3 Answers

You need to use the following notation for the length and the width of the pool: L and w

Since the problem provides the information that the pool is 6 feet longer than its wide, hence L = 6+w.

You need to evaluate the area of the pool:

A_p = L*w => A_p = (6+w)*w

The problem provides the information that the pool is surounded by a walk of 4 ft wide, hence the dimensions of pool and walk are: length = 8+L and width = 8+w

A_w = (8+L)(8+w) => A_w = (8+6+w)(8+w) =>A_w = (14+w)(8+w)

Since area of pool and walk is 272 sq feet more than area of pool, then you may write such that:

(14+w)(8+w) = 272+(6+w)*w

You need to open the brackets such that:

112 + 22w + w^2 - 272 - 6w - w^2 = 0

16w-160 = 0 => 16w = 160 => w = 10 feet

L = 6+w => L = 10+ 6=> L = 16 feet

**Hence, evaluating the dimensions of pool yields the length of 16 feet and the width of 10 feet.**

Let the shorter side of the pool = x ft. yhen longer side = x+6

Area of the pool = x(x+6)= x^2 +6x

Given, pool is surrounded by a walk way of width = 4 ft.

shorter side of the poo along with the walk way = x+8

longer side of the poo along with the walk way = x+6+8 = x+14

Area of the pool including the walk way = (x+8)(x+14)=x^2+8x+14x+112 =x^2 +22x +112

It is given area of the pool & walk is 272 ft. more than area of the pool

Thus, ( x^2 +22x +112) -(x^2 +6x) = 272

=> 16x +112 = 272

=> 16x = 160

=> x = 10 i.e shorter side(x)= 10 ft, Therefore longer side = x+6= 16 ft.

the dimensions of the pool : 10 ft. and 16 ft. <---Answer

N.B --> this is not a discussion at all , This should be posted in Question & ans section.

Let the shorter side of the pool be x feet so that its length is x+6. The walk way outer dimensions would be x+8 feet wide and x+14 feet long, as the walk way width is 4 feet.

The area of walk-way along with pool = (x+8)(x+14) = x^2+22x+112 sq.ft.

The area of pool = x(x+6) = x^2+6x sq.ft.

As the area of the pool and the walk is 272 square feet more than the area of the pool, therefore:

x^2+22x+112 - (x^2+6x) = 272

16x = 272-224 = 160 feet

x = 10 feet

Therefore width = 10 feet and length = 10+6 = 16 feet

**The dimensions of pool are 10 feet wide and 16 feet long.**

P.S.

*The question should have been put in Q&A section rather than as a discussion topic.*