The dimensions of rectangle can anyone from the set [tex]\{1 \times 4,2 \times 5,3 \times 6,4 \times 7,5 \times 8,6 \times 9,7 \times 10,8 \times 11,9 \times 12,10 \times 13,11 \times 14\}[/tex]
Solution:
Given that , The length of a rectangle is 3 inches greater than its width.
The perimeter is less than 54 inches.
Now, let the width of the rectangle be "n" inches. Then, the length of the rectangle will be n + 3 inches.
perimeter = 2(length + width)
Substituting the values, we get,
perimeter = 2(n + 3 + n) = 2 (2n + 3)
perimeter = 4n + 6
Now, we are given that, perimeter is less than 54
Then, perimeter < 54
4n + 6 < 54
4n < 54 – 6
4n < 48
n < 12
So, that, the width of the rectangle can be any value less than 12 inches.
Such that the values of width of rectangle are from 1 to 11
And corresponding length of rectangle are 1 + 3 to 11 +3 ⇒ 4 to 14
Thus Dimensions of rectangle can anyone from the set [tex]\{1 \times 4,2 \times 5,3 \times 6,4 \times 7,5 \times 8,6 \times 9,7 \times 10,8 \times 11,9 \times 12,10 \times 13,11 \times 14\}[/tex]