The area of a rectangular window is to be 306 square centimeters. If the length exceeds the width by 1 centimeter, what are the dimensions?

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- Jan 11th 2007, 01:56 PMsymmetryWindow Dimensions
The area of a rectangular window is to be 306 square centimeters. If the length exceeds the width by 1 centimeter, what are the dimensions?

- Jan 11th 2007, 11:20 PMmachi4velli
The area of a rectangle is length (l) * width (w)

l = w + 1

(length is 1 more than width)

We can now write the problem as this:

306 = w(w + 1)

306 = w^2 + w

0 = w^2 + w - 306

We can use the quadratic formula to find the solutions:

0 = (-b +- sqrt(b^2 - 4ac))/2a

The coefficient of w^2 is 1, so a = 1; the coefficient of w is 1, so b = 1; and c = -306.

0 = (-1 +- sqrt(1^2 - 4(1)(-306)))/2(1)

0 = (-1 +- sqrt(1 + 1225))/2

0 = (-1 +- 35)/2

This gives us 17 for one solution:

(-1 + 35)/2

34/2

17

and 18 for the other solution:

(-1 - 35)/2

-36/2

-18

Since the width cannot be negative, the width must be 17.

l = w + 1

l = 17 + 1

l = 18

The length is 18 cm and the width is 17 cm. - Jan 11th 2007, 11:42 PMCaptainBlack
If the length is x cm, then the width is x-1, so the area of the window is:

x(x-1)=306.

Solve this with the quadratic formula and you find x=-17 or x=18 are the

roots the first is a non-physical result and so we discard it, leaving the length

as 18 cm, and the width as 17 cm.

Check 17*18=306, so the result is OK

RonL - Jan 12th 2007, 03:32 AMsymmetryok
Thank you both for your great help.