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
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.
area = 144 square units perimeter = 48 units
No rectangle can have equal perimeter and length.
There would be an infinite number of rectangles possible
thare is only 1 differint rectangles
No
2l+2w
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