Re: unknown's question at Yahoo! Answers regarding the perimeter of a rectangle
Hello unknown,
Let's let one pair of parallel sides, the smaller pair have length $2n-1$ and the other pair, the larger pair, have length $2n+1$, where $n$ is a natural number. We know they are both odd, because $2n$ must be even, and so adding/subtracting $1$ from an even number results in an odd number. We know they are consecutive odd numbers because their difference is $2$:
$$(2n+1)-(2n-1)=2n+1-2n+1=2$$
So, let's draw a diagram of our rectangle:
View attachment 1352
We see the perimeter $P$, which is the sum of the lengths of the four sides is:
$$P=(2n+1)+(2n-1)+(2n+1)+(2n-1)=8n$$
We are told that this perimeter must be at least 35 (measures in cm), so we may write:
$$8n\ge35$$
Dividing through by $8$, we find:
$$n\ge\frac{35}{8}=4+\frac{3}{8}$$
Since $n$ is a natural number (a positive integer), we may then conclude we must have:
$$n=5$$
And so the two larger sides have length:
$$2(5)+1=11$$
And the smaller sides have length:
$$2(5)-1=9$$
This gives us a perimeter of $40\text{ cm}$. If we take the next smallest pair of consecutive integers, namely $7$ and $9$, we find the perimeter would be $32\text{ cm}$.
