Can a rectangle with a area of 42 have a perimeter of 48?

Answer 1

We can solve this by taking our basic equations for the perimeter and area of a rectangle:

a = lw

p = 2(l + w)

And then plugging the given values into those:

42 = lw

48 = 2(l + w)

Now we can solve one of them for either variable. We'll go with solving the first one for l:

l = 42/w

And then we can plug that into the other one:

48 = 2(42/w + w)

And solve for w:

48 = 2(42/w + w)

48 = 84/w + 2w

48w = 84 + 2w2

2w2 - 48w + 84 = 0

w2 - 24w + 42 = 0

w2 - 24w + 144 = 102

(w - 12)2 = 102

w - 12 = ± √102

w = 12 ± √102

So yes, that is indeed possible, and it's length and width will be 12 - √102 and 12 + √102 (or approximately 1.9005 by 22.0995).