SOLUTION: I need your help on the following word problem: A rectangle is 15cm wide and 18cm long. If both dimensions are decreased by the same amount, the area of the new rectangle is 116cm

Algebra ->  Calculators  -> Pythagorean Theorem Solver, Cartoon Lesson, and practice -> SOLUTION: I need your help on the following word problem: A rectangle is 15cm wide and 18cm long. If both dimensions are decreased by the same amount, the area of the new rectangle is 116cm      Log On


   

Question 171289: I need your help on the following word problem:
A rectangle is 15cm wide and 18cm long. If both dimensions are decreased by the same amount, the area of the new rectangle is 116cm squared less than the area of the original. Find the dimensions of the new rectangle.
Thank You
Found 2 solutions by gonzo, solver91311:
Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
width of rectangle equals 15 centimeters.
length of rectangle equals 18 centimeters.
-----
let L = length
let W = width
let A = area.
-----
area of triangle = length times width:
A = L * W
-----
If both dimensions are decreased by the same amount, the area of the new rectangle is 116cm squared less than the area of the original.
-----
let x be the amount the length and the width are decreased by.
your formula to solve would be:
A - 116 = (L-x) * (W-x)
since A is the original area and it is equal to L * W, you can substitute in the equation to get:
(L*W) - 116 = (L-x)*(W-x)
since you know what L is and you know what W is you can substitute for them in the equation:
(18*15) - 116 = (18-x)*(15-x)
simplify:
270 - 116 = 270 - 18*x - 15*x + x^2
subtract 270 from both sides and it simplify further:
- 116 = -33*x + x^2
add 116 to both sides and rearrange to the standard order exponents (highest order first):
x^2 - 33*x + 116 = 0
solve the quadratic equation of:
x^2 - 33*x + 116 = 0
-----
i didn't see it right away, so i used the quadratic formula of:
x+=+%28%28-b%29%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
to solve the quadratic equation.
-----
the standard form of the quadratic equation is:
a%2Ax%5E2+%2B+b%2Ax+%2B+c+=+0
a is the coefficient of x^2
b is the coefficient of x
c is the constant term
-----
in your equation,
a = 1
b = -33
c = 116
-----
the factors came out to be:
(x-29) * (x-4) = 0
x = 29
or:
x = 4
-----
to see which of these answers is correct, we need to substitute them in the original equation.
-----
the original equation is:
(18*15) - 116 = (18-x)*(15-x)
substitute 29 for x:
(18*15) - 116 = (18-29)*(15-29)
simplify:
270 - 116 = (-11)*(-14)
154 = 154
x = 29 works.
-----
substitute 4 for x:
(18*15) - 116 = (18-4)*(15-4)
simplify:
270 - 116 = (14)*(11)
154 = 154
x = 4 works as well.
-----
your answer is
x = 4 centimeters
or
x = 29 centimeters
-----

Answer by solver91311(16897) About Me  (Show Source):
You can put this solution on YOUR website!
If the original dimensions are 15 by 18 and they are both reduced by the same amount, then the new dimensions are x and x+%2B+3. The original area was 15 times 18 or 270 square centimeters. 116 square centimeters less than 270 is 154 square centimeters which is the area of the new rectangle.

So:
x%28x%2B3%29=154

x%5E2%2B3x-154=0

Since -11%2A14=-154 and -11%2B14=3, the quadratic factors to:

%28x-11%29%28x%2B14%29=0, so

x=11 or x=-14

-14 is absurd because you can't have a negative dimension, so exclude this root as extraneous. That means that the new width of the rectangle is 11, and the new length is 3 more than that or 14.

Check: 11 * 14 = 154, 154 + 116 = 270, 15 - 4 = 11, and 18 - 4 = 14. So all the original problem parameters are satisfied. Answer checks.