[ Area of a quadrilateral ] [ Quadrilateral (inequalities) ]

The diagonals of a convex quadrilateral are equal to $d_{1}$ and $d_{2}$. What is the maximum value that its area can have?
The area of a convex quadrilateral is equal to half the product of its diagonals by the sine of the angle between them.

## Solution

Let $S$ be the area of the given quadrilateral, $\alpha$ be the angle between its diagonals. Then

$$
S=\frac{1}{2} d_{1} d_{2} \sin \alpha \leqslant \frac{1}{2} d_{1} d_{2}
$$

Equality is achieved when the diagonals are perpendicular to each other.

![](https://cdn.mathpix.com/cropped/2024_05_06_c271881b12104a3b803eg-37.jpg?height=369&width=689&top_left_y=2554&top_left_x=679)
[ Unusual constructions $]
Difficulty: 3
Problem