SOLUTION: A 60-Ft by 80-Ft parking lot is torn up to install a sidewalk of unirom width around it's perimeter. The new area of the parking lot is two-thirds of the old area. How wide is the

Question 249487: A 60-Ft by 80-Ft parking lot is torn up to install a sidewalk of unirom width around it's perimeter. The new area of the parking lot is two-thirds of the old area. How wide is the sidewalk? Use the five step method to solve using algebra only.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the old parking lot is 60 feet by 80 feet.

let the length be 80 feet.

let the width be 60 feet.

the original area of the parking lot is 60 * 80 = 4800 square feet.

the new parking lot is going to be 2/3 * 4800 = 3200 square feet.

let x = the width of the sidewalk

this means that:

(60-x) * (80-x) = 3200

simplify by multiplying out the factors to get:

4800 - 60*x - 80*x + x^2 = 3200

simplify by combining like terms to get:

4800 - 140*x + x^2 = 3200

subtract 3200 from both sides of the equation to get:

4800 - 140*x + x^2 - 3200 = 0

simplify by combining like terms to get:

1600 - 140*x + x^2 = 0 which is the same as:

x^2 - 140*x + 1600 = 0

the standard form of a quadratic equation is:

ax^2 + bx + c = 0

in your equation:

a = 1
b = -140
c = 1600

use the quadratic formula to solve this since you can't find the factors easily by using the factor method.

the quadratic formula is:

you will get:

x = 127.4456265 or x = 12.55437353

substitute in your original equation and you will see that both answers are valid solutions to the equation.

x = 127.4456265, however, is too large to be considered since it is greater than 80 or 60.

your answer has to be:

x = 12.55437353

the dimensions of the new parking lot will be:

60 - 12.55437353 = 47.44562647 feet.

80 - 12.55437353 = 67.44562647 feet.

47.44562647 * 67.44562647 = 3200 square feet.

the answers are confirmed as good.

the answer to your problem is:

the width of the sidewalk is 12.55437353 feet.

RELATED QUESTIONS
A 60-ft. by 80-ft. parking lot is torn up to install a sidewalk of uniform width around... (answered by josgarithmetic,MathTherapy)

I am not sure how to set this up so I can solve it. Your help is greatly appreciated. (answered by ptaylor)

A 50 foot by 60 foot parking lot is torn up to install a sidewalk of uniform width around (answered by ankor@dixie-net.com)

A parking lot is 75 feet wide by 60 feet long. It is being torn up to install a sidewalk... (answered by ankor@dixie-net.com)

A rectangular garden is 60 ft by 80 ft. Part of the garden is torn up to install a... (answered by ankor@dixie-net.com)

A rectangular lawn measures 60ft. by 80ft. Part of the lawn is torn up to install a... (answered by josgarithmetic,ankor@dixie-net.com)

A rectangular lawn measures 60ft. by 80ft. Part of the lawn is torn up to install a... (answered by josgarithmetic)

a rectangular lawn is 80 ft by 60ft. part of the lawn is torn up to install a sidewalk... (answered by stanbon)

A rectangular field measures 80 ft by 60ft. Part of the lawn is torn up to install a walk (answered by ankor@dixie-net.com)