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).
28 inches
maybe
You might want to investigate the rectangle that measures [ 6 by 8 ].
7 x 17
Anything greater than 27.713cm (4 x √48).
If the rectangle is a square, the perimeter is 48 cm. If not, there are a lot of possibilities.
2*24
28 inches
maybe
You might want to investigate the rectangle that measures [ 6 by 8 ].
7 x 17
What are the dimensions of a rectangle that has a perimeter of 56 units and an area of 96 square units
Anything greater than 27.713cm (4 x √48).
If the length of a rectangle is twice its width and it has a perimeter of 48, then the rectangle is 16 in length and 8 in width.
The maximum area is attained when the rectangle is, in fact, a square. Since the perimeter = 48 feet, the maximum length for a square = 48/4 = 12 feet. So max area = 122 = 144 square feet.
I thought circumference is only with circles Absolutely correct, but even if you consider the perimeter, which is conceptually similar, there is no answer. A rectangle with an area of 24 could be 4*6 with a perimeter of 20 or it could be 1*24 with a perimeter of 50 or 0.5*48 with a perimeter of 97 etc
No, 12x4 = 48 sq meters. 13x3=39 sq meters.