The length of a picture frame is 3 in. greater than the width. The perimeter is less than 52 in. How do you find the dimensions of the frame?

Question

Aurora Ayala · Accepted Answer

We can at once replace #L=W+3#

#P=2xxL+2xxW=2xx(W+3)+2xxW# #P=2W+6+2W=4W+6#

Now since #P<52#, we get: #4W+6<52# subtracting 6: #4W<52->W<13#

Conclusion: Width is less than 13 inches Length is less than 16 inches

Note : There could not be just any combination of #L<16andW<13# as #L=W+3# still holds. (so #L=15, W=10# is not allowed)

Noah Atkinson · Answer

undefined Let's denote the width of the picture frame as $ W $ inches. Since the length of the frame is 3 inches greater than the width, we can represent the length as $ W + 3 $ inches.

The perimeter of a rectangle (picture frame) is given by the formula $ P = 2L + 2W $, where $ P $ is the perimeter, $ L $ is the length, and $ W $ is the width.

Given that the perimeter is less than 52 inches, we have the inequality:

$ 2(W + 3) + 2W < 52 $

Simplifying and solving for $ W $, we get:

$ 2W + 6 + 2W < 52 $

$ 4W + 6 < 52 $

$ 4W < 46 $

$ W < \frac{46}{4} $

$ W < 11.5 $

So, the width of the frame must be less than 11.5 inches. Since the width cannot be negative or zero, we consider $ W $ to be an integer. The possible values for $ W $ would be 1, 2, 3, ..., 11.

Next, we can find the corresponding lengths for each possible width $ W $ by using $ L = W + 3 $.

For example, if $ W = 1 $, then $ L = 1 + 3 = 4 $. This would give us a perimeter of $ 2(4) + 2(1) = 8 + 2 = 10 $ inches.

Similarly, we can calculate the perimeters for $ W = 2, 3, \ldots, 11 $ to see which combinations result in a perimeter less than 52 inches. The dimensions of the frame that satisfy the given conditions would be the ones where the perimeter is less than 52 inches.