52. The median line of a trapezoid divides it into two quadrilaterals. The difference in the perimeters of these two quadrilaterals is 24, and the ratio of their areas is $\frac{20}{17}$. If the height of the trapezoid is 2, then the area of this trapezoid is $\qquad$ _.
Answer: 148
Solution: Let the upper base of the trapezoid be $\mathrm{a}$, and the lower base be $\mathrm{b}$, then the length of the midline is
$$
\frac{a+b}{2} 。
$$

The midline of the trapezoid bisects the two legs, and the two quadrilaterals it divides the trapezoid into are both trapezoids,
so the difference in the perimeters of these two parts is

the ratio of the areas is
$$
b-a=24,
$$
$$
\frac{b+\frac{a+b}{2}}{a+\frac{a+b}{2}}=\frac{20}{17} \text {, }
$$

we get
$$
31 b=43 a \text {, }
$$

From (1) and (2), we can solve for
$$
a=62, b=86 \text {, }
$$

Therefore, the area of the trapezoid is
$$
(62+86) \times 2 \div 2=148 \text {. }
$$