The parabola $y = x^2 + 16$ intersects the $x$-axis at points A and B. Further, the parabola intersects the line $y = 7$ at points M and N. Then the area of the quadrilateral with vertices A, B, M and N is

Assuming the equation of the parabola is $y = -x^2 + 16$ (since $y = x^2 + 16$ does not intersect the x-axis at real points):

To find the intersection points A and B with the x-axis, substitute $y = 0$:

$-x^2 + 16 = 0 \Rightarrow x = \pm 4$

The coordinates of A and B are $(-4, 0)$ and $(4, 0)$. The length of the side $AB = 8$.

To find the intersection points M and N with the line $y = 7$, substitute $y = 7$:

$-x^2 + 16 = 7 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$

The coordinates of M and N are $(-3, 7)$ and $(3, 7)$. The length of the side $MN = 6$.

The quadrilateral ABMN is a trapezium with parallel sides AB and MN. The height of the trapezium is the perpendicular distance between the lines $y = 0$ and $y = 7$, which is $7$.

Area of the trapezium $= \dfrac{1}{2} \times (AB + MN) \times h$

Area $= \dfrac{1}{2} \times (8 + 6) \times 7 = 49$

Answer: $49$