$\left[\begin{array}{l}{[\text { Midline of the trapezoid }]} \\ {[\quad \text { Area of the trapezoid }]}\end{array}\right]$

The product of the midline of the trapezoid and the segment connecting the midpoints of its diagonals is 25. Find the area of the trapezoid, if its height is three times the difference of the bases.
It is known that the midline of a trapezoid is equal to the half-sum of the bases, and the segment connecting the midpoints of the diagonals is the half-difference of the bases. Let $a$ and $b$ be the bases of the trapezoid ($a > b$), $h$ be its height, and $S$ be its area. According to the problem,

$$
\frac{a+b}{2} \cdot \frac{a-b}{2}=25, \quad h=3(a-b)
$$

Therefore,

$$
S=\frac{a+b}{2} \cdot h=\frac{a+b}{2} \cdot 3(a-b)=\frac{3}{2}(a+b)(a-b)=\frac{3}{2} \cdot 25 \cdot 4=150 .
$$

## Answer

150.00