10.237. The diagonals of a quadrilateral are equal, and the lengths of its midlines are $p$ and $q$. Find the area of the quadrilateral.
Solution.

Since $K L, L M, N M, K N$ are the midlines of the corresponding triangles (Fig. 10.44), then $K L=N M=\frac{1}{2} A C, K N=L M=\frac{1}{2} B D$. By the condition $A C=B D \Rightarrow K L M N-$ is a rhombus, the area of which is equal to $\frac{1}{2} p q$. Let the required area be $S$. Then $S=S_{\triangle A B D}+S_{\triangle B C D}$. But $S_{\triangle A K N}=\frac{1}{4} S_{\triangle A B D}$;

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0739.jpg?height=419&width=581&top_left_y=65&top_left_x=81)

Fig. 10.44

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0739.jpg?height=440&width=650&top_left_y=151&top_left_x=610)

Fig. 10.45

Similarly, $S_{\triangle C M L}=\frac{1}{4} S_{\triangle B C D}$, so $S_{\triangle A K N}+S_{\triangle C M L}=\frac{1}{4} S$. Similarly, we obtain that $S_{\triangle K B L}+S_{\triangle M D N}=\frac{1}{4} S$. Thus, the area outside the rhombus is $\frac{1}{2} S$. Therefore, the area of the rhombus is also $\frac{1}{2} S$, from which $S=p q$.

Answer: $p q$.