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The length of a rectangle is twice its width. The perimeter is 60ft.. Find the area. Explain in complete sentences the steps necessary to find the area.

I need to know the answer to this question. The length of a rectangle is twice its width. The perimeter is 60ft. What is the area. Please explain in complete sentences the steps necessary to find the area.

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Yossi Y. | Electronic Engineering Background and Teaching/Tutoring ExperienceElectronic Engineering Background and Te...
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Set the following: Length = L    Width = W

The length is twice the width:    L = 2*W     or:  W = L/2

Perimeter: P = 60ft = 2*L + 2*W

Substituting L for W in the perimeter equation gives:

P = 60ft = 2*L + 2*(L/2) = 2*L + L = 3L

L = 60ft/3 = 20ft    W=L/2=20ft/2=10ft

Verify result:  Perimeter= P = 60ft = 2L + 2W = 2*20ft +2*10ft = 40ft+20ft = 60ft

The area of the rectangular is: Area = A = Length*Width = L*W = 20ft*10ft = 200 (ft)^2

 

 

Michael B. | Seasoned and experienced tutor with extensive science backgroundSeasoned and experienced tutor with exte...
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Since the statement, "the length is twice its width" puts one dimension in terms of the other it's easiest to use "w" as the variable.

So,

w = width

2w = length

The perimeter is the sum off all the sides of the rectangle

P = w + w + l + l = 2w + 2l

Using our terms and the value of 60, substitute them into the equation.

60 = 2w + 2(2w)

60 = 6w

w = 10 ft

l = 2w = 20 ft

Checking our solution:

60 ft = 10 ft + 10 ft + 20 ft + 20 ft

60 ft = 60 ft 

Now onto the area, the area of a rectangle is the product of its dimensions and is given by the formula

Area = l*w

substituting our values in from above

A = (20 ft)*(10 ft) = 200 ft2