[ Area of a Quadrilateral ]

The diagonals of a quadrilateral are equal to $a$, and the sum of its midlines is $b$ (midlines connect the midpoints of opposite sides). Calculate the area of the quadrilateral.
Let's connect the midpoints of the sides of the given quadrilateral $E, K, T, P$ (see figure). The resulting quadrilateral $EKT P$ is a rhombus, because its sides are equal to half the diagonals as the midlines of triangles $A B C, A D C, B C D, A B D$, and the diagonals of the given quadrilateral are equal to $a$ by condition. Further, the midlines of the specified triangles cut off triangles from them with an area that is 4 times smaller than the areas of these triangles, i.e., $S_{B E K}=1 / 4 S_{B A C}, S_{P T D}=1 / 4 S_{A C D}, S_{K C T}=1 / 4 S_{D B C}, S_{A E P}=1 / 4 S_{A B D} \cdot$ Adding the areas of the first two triangles, and then the second two, we get:

$$
\begin{aligned}
& S_{B E K}+S_{P T D}=1 / 4\left(S_{A B C}+S_{A C D}\right)=1 / 4 S_{A B C D} \\
& S_{K C T^{+}}+S_{A E P}=1 / 4\left(S_{D B C}+S_{A B D}\right)=1 / 4 S_{A B C D}
\end{aligned}
$$

The sum of the areas of all four triangles that complement the rhombus is half the area of the quadrilateral. Therefore, the area of the given quadrilateral is twice the area of the rhombus. The area of the rhombus is half the product of its diagonals, and the area of the given quadrilateral is the product of the diagonals of the rhombus, which are the midlines of the given quadrilateral. If we denote the midlines by $x$ and $y$, then from triangle, for example, $O E K$ we find that $x^{2}+y^{2}=a^{2},(x+y)^{2}=a^{2}+2 x y, x y=\left((x+y)^{2}-a^{2}\right) / 2=\left(b^{2}-a^{2}\right) / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6b47f7415227382cfc43g-18.jpg?height=555&width=666&top_left_y=1189&top_left_x=707)

## Answer

$\left(b^{2}-a^{2}\right) / 2$