7. Given that the upper base, height, and lower base of a trapezoid are three consecutive positive integers, and these three numbers make the value of the polynomial $x^{3}-30 x^{2}+a x$ (where $a$ is a constant) also three consecutive positive integers in the same order. Then the area of this trapezoid is $\qquad$
7. 100 .

Let the upper base, height, and lower base be $n-1$, $n$, and $n+1$ respectively, and the other three consecutive integers be $m-1$, $m$, and $m+1$. Then,
$$
\begin{array}{l}
(n-1)^{3}-30(n-1)^{2}+a(n-1)=m-1, \\
n^{3}-30 n^{2}+a n=m, \\
(n+1)^{3}-30(n+1)^{2}+a(n+1)=m+1 . \\
\text { (1) }+(3)-2 \times(2) \text { gives } \\
6 n-60=0 \Rightarrow n=10 .
\end{array}
$$

Therefore, the area of the trapezoid is
$$
\frac{1}{2}(9+11) \times 10=100 \text {. }
$$