To solve the problem of drawing three different rectangles that each have a perimeter of 22 cm, we should start by understanding the formula for the perimeter of a rectangle. The perimeter [tex]\( P \)[/tex] of a rectangle with length [tex]\( a \)[/tex] and width [tex]\( b \)[/tex] is given by:
[tex]\[ P = 2 \times (a + b) \][/tex]
Given that the perimeter is 22 cm, we can set up the equation:
[tex]\[ 2 \times (a + b) = 22 \][/tex]
By dividing both sides of the equation by 2, we get:
[tex]\[ a + b = 11 \][/tex]
Our task now is to find three pairs [tex]\((a, b)\)[/tex] that satisfy this equation. Additionally, we should choose pairs where the length [tex]\( a \)[/tex] and width [tex]\( b \)[/tex] are different to ensure we have distinct rectangles.
Here are the steps to find three such pairs:
1. Choose [tex]\( a = 1 \)[/tex]:
[tex]\[ a + b = 11 \][/tex]
[tex]\[ 1 + b = 11 \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ b = 10 \][/tex]
Therefore, one pair is [tex]\( (a, b) = (1, 10) \)[/tex].
2. Choose [tex]\( a = 2 \)[/tex]:
[tex]\[ a + b = 11 \][/tex]
[tex]\[ 2 + b = 11 \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ b = 9 \][/tex]
Therefore, another pair is [tex]\( (a, b) = (2, 9) \)[/tex].
3. Choose [tex]\( a = 3 \)[/tex]:
[tex]\[ a + b = 11 \][/tex]
[tex]\[ 3 + b = 11 \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ b = 8 \][/tex]
Therefore, a third pair is [tex]\( (a, b) = (3, 8) \)[/tex].
Thus, the three different rectangles with a perimeter of 22 cm are:
1. Rectangle with dimensions 1 cm by 10 cm
2. Rectangle with dimensions 2 cm by 9 cm
3. Rectangle with dimensions 3 cm by 8 cm
You can now draw these three rectangles on paper based on the dimensions provided above.
