Check out some similar questions!

holly_arato · Feb 22, 2013, 02:51 PM

The original length of the rectangle is 3 more than 3x it's width. If the Length and Width are both increased by 4cm the resultant area is 198 cm^2. What are the original dimensions?

mendozaulises78 · Feb 22, 2013, 04:00 PM

Equation 1. L = 3W+3;
Equation 2. Area = L*W;

Equation 3. Area = (L+4)*(W+4)=198

We substitute Equation 1 in 3

((3W+3)4)*(W+4)=198
(3W+7)*(W+4) = 198
3W^2+19W+28=198
3W^2+19W-170=0;

3W^2+19W-170=0 --> (ax^2+bx+c) --> a=3, b=19, c = -170
To solve:
x = (-b +/- sqrt(b^2-4aC) )/2a

W = (-19 +/- sqrt((19^2)-4(3)(-170))/2(3)
W= (-19+/- sqrt(361+2040))/6
W = (-19 +/- sqrt(2401))/6
W = (-19+/- 49)/6
W1 = (-19-49)/6
W2=(-19+49)/6

W1 = -68/6 = -11.83 --> We discard this because it is negative
W2 = 30/6 = 5. --> This is the value of the width

So
W = 5
replacing in equation 1.

L = 3W+3
L = 3(5)+3=18

If we add 4 to each length and widht we have an area of

(5+4)*(18+4) = 9*22=198

So the width is 5 cm, the length is 18 cm.