If the lengths of the sides of a rectangle are integers, and if one side is 2 more than another, then the perimeter could equal
A. 1998
B. 1999
C. 2000
D. 2002
I started this, but I'm not sure what they're asking and I don't know how to solve it.
P=2L+2(L-2)=4L-4
so which of the following is an integer?
(1998/4-1)
(1999/4-1)
(2000/4-1)
(2002/4 -1)
Esperanza makes a rectangle with a price of string. She says the perimeter of her rectangle is 33 centimeters. Explain how it's possible for her rectangle to have an odd perimeter
This is so hard and this is not even a answer
To solve this problem, we need to understand that the perimeter of a rectangle is the sum of all its side lengths. In this case, we have a rectangle where one side is 2 more than the other side. Let's call the shorter side "x" and the longer side "x + 2".
To find the perimeter, we need to add up all the side lengths:
Perimeter = x + x + (x + 2) + (x + 2)
Simplifying the equation, we get:
Perimeter = 4x + 4
Now, we need to determine what values of "x" would result in an integer perimeter. We know that the lengths of the sides are integers, so "x" must also be an integer.
To check the answer choices, we can substitute the values of "x" into the equation and see if the resulting perimeter is a match. Let's go through each answer choice:
A. Perimeter = 4(2) + 4 = 12 (not equal to 1998)
B. Perimeter = 4(2) + 4 = 12 (not equal to 1999)
C. Perimeter = 4(2) + 4 = 12 (not equal to 2000)
D. Perimeter = 4(2) + 4 = 12 (not equal to 2002)
From this analysis, we can see that none of the answer choices result in a valid perimeter. Therefore, the correct answer is none of the above.
In conclusion, the perimeter cannot equal any of the given answer choices (A, B, C, or D) if the lengths of the sides of the rectangle are integers and one side is 2 more than the other side.