P=2(L+w) L=length W=width
perimeter = 2 (b+h) = 20 there are an infinite number of rectangles that meet the requirement
Yes, it can because a 3 by 6 rectangle has the perimeter of 18 and has the area of 18! :)
Yes. Say there are two rectangles, both with perimeter of 20. One of the rectangles is a 2 by 8 rectangle. The area of this rectangle is 2 x 8 which is 16. The other rectangle is a 4 by 6 rectangle. It has an area of 4 x 6 which is 24.
This browser is hopeless for drawing but consider the following two rectangles: a*b and (a+1)*(b-1). Their perimeter will be 2a+2b but unless a = b-1, their area will be different.
To find all the rectangles with a perimeter of 26 cm, we can use the formula for the perimeter of a rectangle: P = 2(l + w), where P is the perimeter, l is the length, and w is the width. Since the perimeter is given as 26 cm, we have 26 = 2(l + w). We can then list all the possible combinations of length and width that satisfy this equation, such as (length 10 cm, width 3 cm), (length 8 cm, width 5 cm), and so on.
There is an infinite number that can have that perimeter
No, not all rectangles have even perimeters. The perimeter of a rectangle is calculated using the formula ( P = 2(length + width) ). If either the length or width is an odd number, their sum can be odd, resulting in an odd perimeter when multiplied by 2. Therefore, a rectangle can have an odd perimeter if one or both dimensions are odd.
10cm by 10cm (perimeter=40cm), 5cm by 20cm (perimeter=50cm), 50cm by 2cm (perimeter=104cm), 100cm by 1cm (perimeter=202cm). All of these rectangles' areas are 100cm2
they dont
perimeter = 2 (b+h) = 20 there are an infinite number of rectangles that meet the requirement
The perimeter of a rectangle is the sum of its four sides. Add the sides for both rectangles, then compare the results.
No, two rectangles do not have to be congruent if they have the same perimeter. Rectangles can have the same perimeter while differing in their length and width. For example, a rectangle with dimensions 4x6 has the same perimeter (20 units) as a rectangle with dimensions 5x5, but they are not congruent since their shapes and sizes differ.
area = 144 square units perimeter = 48 units
No rectangle can have equal perimeter and length.
thare is only 1 differint rectangles
There would be an infinite number of rectangles possible
No