Thread: Window 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?

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.

Originally Posted by symmetry

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?

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

Thank you both for your great help.