Question

Jamie has a rectangular pool that has a length of 12 feet and a

Answer 1

Answer:

8 feet

Step-by-step explanation:

Given,

Perimeter of a rectangular pool (P) = 40 feet

Length of the pool (l) = 12 feet

Let,

Width of the pool be = w

As we know,

• Perimeter of a rectangle = 2(length + width)

Therefore,

By the problem,

=> 2(l + w) = P

• [On substituting the values of l = 12 and P = 40]

=> 2(12 + w) = 40

• [On multiplying 2 with 12 and w]

=> 24 + 2w = 40

• [On subtracting both sides with 24]

=> 24 - 24 + 2w = 40 - 24

• [On Simplifying]

=> 2w = 16

• [On Dividing both sides with 2]

=> $\frac{2w}{2}$ = $\frac{16}{2}$

• [On Simplifying]

=> w = 8

Hence,

The required width of the pool is 8 feet. (Ans)

Answer 2

Answer:

The Width of the pool would be 8 feet/ft. .

Step-by-step explanation:

According to the Question Given:

Perimeter = 40 ft/feet

Length of the pool = 12 ft/feet

To Find:

The width/breadth of the pool

Solution:

We know that,

$\boxed{\tt \: Perimeter \: of \: Rectangle = 2(l + b)}$

So Put their values accordingly:

• Perimeter of The Rectangle = 40
• Length[L] = 12

$\longrightarrow \tt \: 40 = 2(12 + b)$

We got an equation.By this method we can easily find the breadth/width of the pool.

Solve this equation:

$\longrightarrow \tt40 = 2b + 24$

Flip the equation:

$\longrightarrow \tt2b + 24 = 40$

• Transpose 24 to the RHS[remember to change its sign]:

$\longrightarrow \tt2b = 40 - 24$

• Simplify:

$\longrightarrow \tt2b = 16$

Divide both sides by 2:

$\tt\longrightarrow \cfrac{2b}{2} = \cfrac{16}{2}$

• Use Cancellation method and cancel LHS and RHS:

$\tt\longrightarrow \cfrac{ \cancel2 {}^{1} b}{ \cancel2} = \cfrac{ \cancel{16} {}^{8} }{ \cancel2}$

$\longrightarrow \tt1b = 8$

$\longrightarrow \tt{b} =\boxed{\tt 8 \: feet}$

Hence, the breadth/width of the pool would be 8 ft./feet .

$\rule{225pt}{2pt}$

I hope this helps!