7. Given a triangle with sides as three consecutive natural numbers, the largest angle is twice the smallest angle. Then the perimeter of the triangle is $\qquad$
II, 7.15.
Let the three sides of $\triangle ABC$ be $AB=n+1, BC=n, AC=n-1$, and the angle opposite to side $AC$ be $\theta$, and the angle opposite to side $AB$ be $2\theta$.
According to the Law of Sines and the Law of Cosines, we have
$$
\begin{array}{l}
\frac{n-1}{\sin \theta}=\frac{n+1}{\sin 2 \theta} \Rightarrow \frac{n+1}{n-1}=2 \cos \theta, \\
\cos \theta=\frac{(n+1)^{2}+n^{2}-(n-1)^{2}}{2 n(n+1)} \\
=\frac{n+4}{2(n+1)} .
\end{array}
$$

From equations (1) and (2), we get
$$
\frac{n+1}{n-1}=\frac{n+4}{n+1} \Rightarrow n=5 \text {. }
$$

Therefore, the perimeter of the triangle is 15.