SOLUTION: Solving Area and Perimeter Problems: Need to use the RSTUV Method to solve The area of a rectangle is 96 square centimeters. If the width of the rectangle is 4 centimeters le

Algebra ->  Surface-area -> SOLUTION: Solving Area and Perimeter Problems: Need to use the RSTUV Method to solve The area of a rectangle is 96 square centimeters. If the width of the rectangle is 4 centimeters le      Log On


   

Question 480823: Solving Area and Perimeter Problems: Need to use the RSTUV Method to solve
The area of a rectangle is 96 square centimeters. If the width of the rectangle is 4 centimeters less than its length, what are the dimensions of the rctangle?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
RSTUV
1. Read the problem.
2. Select the unknown.
3. Translate into an equation.
4. Use the rules to solve.
5. Verify the result
1.
area of a rectangle is 96 square cm.
width is 4 cm less than its length.
what are dimensions of rectangle.
2.
let L = length
let W = width
let A = area
3.
A = L * W
W = L - 4 cm
Since W = L-4, then the equation for the area of a rectangle becomes:
A = L * (L-4)
4.
A = 96 so equation of:
A = L * (L-4) becomes:
96 = L * (L-4)
perform indicated operations to get:
96 = L^2 - 4L
subtract 96 from both sides of the equation to get:
0 = L^2 - 4L - 96
commute to get:
L^2 - 4L - 96 = 0
this is a quadratic equation.
factor it to get:
(L - 12) * (L + 8) = 0
solve for L to get:
L = 12 or L = -8
L can't be negative so L = 12.
back to original equation of:
A = L * W which becomes:
96 = 12 * W
divide both sides of this equation by 12 to get:
96/12 = W which becomes:
W = 8
5.
the dimensions of the rectangle are:
L = 12
W = 8
the area of the rectangle is 36 square cm.
A = L * W becomes A = 12 * 8 which becomes A = 96
this is the same area we started with so the dimensions of the rectangle look good.
the dimensions are:
L = 12 cm
W = 8 cm