## Task A-4.2.

Let $a, b, c, d$ be positive real numbers such that $a^{2}+b^{2}=c^{2}+d^{2}$.

What is the maximum possible area of the quadrilateral $ABCD$ with side lengths $|AB|=a, |BC|=b, |CD|=c$, and $|DA|=d$?
## Solution.

Diagonal $\overline{A C}$ divides the quadrilateral into two triangles $A B C$ and $C D A$.

For the area of triangle $A B C$, we have $P(A B C)=\frac{1}{2}|A B| \cdot|B C| \cdot \sin \varangle A B C \leqslant \frac{1}{2} a b$.

Similarly, we conclude that $P(C D A) \leqslant \frac{1}{2} c d$.

Since $P(A B C D)=P(A B C)+P(C D A)$, we have shown that the area of quadrilateral $A B C D$ is not greater than $\frac{1}{2}(a b+c d)$.

We still need to show that there exists a quadrilateral $A B C D$ whose area is exactly $\frac{1}{2}(a b+c d)$.

![](https://cdn.mathpix.com/cropped/2024_05_30_11ec2cf7b6def7a4ec70g-18.jpg?height=512&width=902&top_left_y=812&top_left_x=514)

Consider the quadrilateral $A B C D$ where $|A C|=\sqrt{a^{2}+b^{2}}\left(=\sqrt{c^{2}+d^{2}}\right)$. Then triangles $A B C$ and $C D A$ are right triangles, and the area of quadrilateral $A B C D$ is $\frac{1}{2}(a b+c d)$.