Blinkov Yu.A.

In a convex pentagon $A B C D E: \angle A=\angle C=90^{\circ}, A B=A E, B C=C D, A C=1$. Find the area of the pentagon.
Let $A B=A E=a, C B=C D=b, \angle A B C=\alpha$. Then $B E=a \sqrt{2}, B D=b \sqrt{2}, \angle E B D=\alpha-90^{\circ}$ (see the left figure). $S_{A B C D B}=S_{A B B}+S_{C B D}+S_{R B D}=\frac{1}{2} a^{2}+\frac{1}{2} b^{2}+\frac{1}{2} \cdot a \sqrt{2} \cdot b \sqrt{2} \cdot \sin \left(\alpha-90^{\circ}\right)=\frac{1}{2}\left(a^{2}+b^{2}-2 a b \cdot \cos \alpha\right)$. $\quad$ But by the cosine rule $a^{2}+b^{2}-2 a b \cos \alpha=A C^{2}=1$. Therefore, $S_{A B C D E}=0.5$.
![](https://cdn.mathpix.com/cropped/2024_05_06_11215e94233515718ac4g-13.jpg?height=932&width=1012&top_left_y=-2&top_left_x=343)

## Answer

## 0.5 .

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