Geometry involving a diagram and inequalities 
The answer to the following problem is A by the way. Morever how does that come to be?
 A: There are 20 segments of fencing with fence posts 10 feet apart, placed horizontally or vertically: hence we've got at least $$20 \times 10 = 200 + x\;\text{ feet of along the perimeter}$$.
Now, there is an extra fence of length x, which we need to add to complete the computation of the Perimeter P: the diagonal piece, which can be thought of as the diagonal of a $10\times 10$ foot square, and has length (diagonal) of $10\sqrt 2$

$10$ ft $\times 10$ ft square with diagonal $x = 10\sqrt 2$. (Consider the fence piece being the diagonal from top left to bottom right). 
By the Pythagorean Theorem: $$x^2 = 10^2 + 10^2 \implies x = \sqrt{100 + 100} = \sqrt{2\times 100} = \sqrt 2 \times \sqrt {100} = 10\times\sqrt 2 \approx14.1421$$
Hence the perimeter is $$200 + x = 200 + 10 \sqrt2 \approx214.1421 > 210 \text{ feet}$$
And note that by the triangle inequality, the length of the diagonal, $10\sqrt 2 < 10 + 10 = 20$, which is less than the sum of two pieces of fencing of length 10.
So, putting that together, we have $$230 \gt 220 \gt 200+10\sqrt{2} = P \gt 210$$
That leaves us with only option (A) being correct: $P > 210$ feet.
A: There are $20$ sections of fence that run either horizontally or vertically between consecutive fence posts and are therefore each $10$ feet long; that accounts for $20\cdot10=200$ feet of fence. The remaining section of fence runs along the diagonal of a square whose sides are $10$ feet long. The diagonal of a square is longer than a side, so that section is more than $10$ feet long. On the other hand, the diagonal is less than the combined length of two sides, so that section is less than $20$ feet long. That pins down the total length of fence as being strictly between $210$ and $220$ feet, and only answer (A) fits.
In fact you can apply the Pythogorean theorem to two sides and the diagonal of that square to see that if $x$ is the length of the diagonal, then $10^2+10^2=x^2$, so $x^2=200$, and $x=10\sqrt2\approx14.1421$ feet.
