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Can the perimeter of a rectangle be 60 units when its width is 11 units and its length is 24 units? Use the equation [tex]\( P = 2L + 2W \)[/tex], where [tex]\( P \)[/tex] is the perimeter, [tex]\( L \)[/tex] is the length, and [tex]\( W \)[/tex] is the width of the rectangle.

A. No. If the length is 24 units and the width is 11 units, the perimeter would be [tex]\( P = 48 + 22 = 70 \)[/tex] units, not 60.

B. No. If the length is 24 units and the width is 11 units, the perimeter would be [tex]\( P = 24 + 22 = 66 \)[/tex] units, not 60.

C. Yes. If the perimeter is 60 units and the width is 11 units, then [tex]\( P + W \)[/tex] is greater than 48.

D. Yes. If the length is 24 units and the width is 11 units, then [tex]\( P = 2L + 2W = 60 \)[/tex].


Sagot :

Let's analyze the given problem and determine if the perimeter of the rectangle can be 60 units with the given length and width.

Given:
- Length [tex]\( L = 24 \)[/tex] units
- Width [tex]\( W = 11 \)[/tex] units

The formula for the perimeter [tex]\( P \)[/tex] of a rectangle is:
[tex]\[ P = 2L + 2W \][/tex]

Substitute the given length and width into the formula:
[tex]\[ P = 2(24) + 2(11) \][/tex]

Calculate the terms:
[tex]\[ P = 48 + 22 \][/tex]
[tex]\[ P = 70 \][/tex]

Therefore, the perimeter of the rectangle with length 24 units and width 11 units is:
[tex]\[ P = 70 \ \text{units} \][/tex]

Now we check the options:

1. No. If the length is 24 units and the width is 11 units, the perimeter would be [tex]\( P = 48 + 22 = 70 \)[/tex] units, not 60.
- This statement correctly calculates the perimeter as 70 units.

2. No. If the length is 24 units and the width is 11 units, the perimeter would be [tex]\( P = 24 + 22 = 66 \)[/tex] units, not 60.
- This statement incorrectly calculates the perimeter as 66 units, which is incorrect.

3. Yes. If the perimeter is 60 units and the width is 11 units, then [tex]\( P + W \)[/tex] is greater than 48.
- This statement is unrelated to the correct calculation of the perimeter.

4. Yes. If the length is 24 units and the width is 11 units, then [tex]\( P = 2L + 2W = 60 \)[/tex].
- This statement is incorrect because the correct perimeter is 70 units, not 60.

Therefore, the correct answer is:
No. If the length is 24 units and the width is 11 units, the perimeter would be [tex]\( P = 48 + 22 = 70 \)[/tex] units, not 60.