[The area of a triangle does not exceed half the product of two sides] Complex

What is the maximum area that a quadrilateral with side lengths of 1, 4, 7, and 8 can have?
Consider a quadrilateral, two adjacent sides of which are equal to 1 and 8. Then the other two adjacent sides are equal to 4 and 7. If the sides equal to 1 and 8 are opposite, we draw a diagonal and reflect one of the resulting triangles relative to its perpendicular bisector. Then the sides equal to 1 and 8 will be adjacent, and the area of the quadrilateral will not change.

Let \( S \) be the area of the given quadrilateral, \( \alpha \) be the angle between the adjacent sides equal to 1 and 8, and \( \beta \) be the angle between the adjacent sides equal to 4 and 7. Then

\[
S = \frac{1}{2} \cdot 1 \cdot 8 \sin \alpha + \frac{1}{2} \cdot 4 \cdot 7 \sin \beta \leqslant \frac{1}{2} \cdot 1 \cdot 8 + \frac{1}{2} \cdot 4 \cdot 7 = 4 + 14 = 18
\]

Indeed, since \( 1^2 + 8^2 = 4^2 + 7^2 \), the condition of the problem is satisfied by a quadrilateral composed of two right triangles with legs 1, 8 and 4, 7 and a common hypotenuse equal to \( \sqrt{65} \).

## Answer

18.