Maximum cycle in a graph with a path of length $k$ I don't understand why this stands:

Let $G$ be a graph containing a cycle $C$, and assume that $G$ contains a path of length at least $k$ between two vertices of $C$. 
  Then $G$ contains a cycle of length at least $\sqrt{k}$.

Since we can extend the cycle $C$ with the vertices of the path, why don't we get a cycle of length $k+2$? ($2$ being the minimum number of vertices belonging to $C$ between the vertices where $C$ connect to it).
I really don't see where that square root is coming from.
For reference this is exercise $3$ from Chapter $1$ of the Diestel book.

The complete graph on $k+1$ vertices shows why you can't get a cycle of length $k+2$. The following example shows why, if you're looking for a long cycle, the best you can hope for in general is a constant times the square root of $k$:
Let $V(G)=\{v_0,v_1,\dots,v_{4n^2}\}$, $E(G)=\{v_iv_{i+1}:0\le i<4n^2\}\cup\{v_{jn}v_{(j+2)n}:0\le j\le{4n-2}\}$. In $G$ there is a path of length $k=4n^2$, each pair of vertices lies on a cycle, and the longest cycle has length $6n-1=3\sqrt{k}-1$.
