Pentagon $A B C D$ is inscribed in a circle of unit radius. It is known that $A B=\sqrt{2}, \angle A B E=45^{\circ}, \angle E B D=$ $30^{\circ}$ and $B C=C D$. Find the area of the pentagon.
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## Solution

Let $R=1-$ be the radius of the circle. Then

$$
A E=2 R \sin \angle A B E=2 \cdot \frac{\sqrt{2}}{2}=\sqrt{2}
$$

Therefore, triangle $E A B$ is a right triangle, $\angle A=90^{\circ}$. Hence, $B E$ is the diameter of the circle, $B E=2$. Then

$$
\begin{gathered}
\angle B D E=90^{\circ}, B D=B E \cos 30^{\circ}=\sqrt{3} \\
\angle D C B=\frac{\cup D A B}{2}=\frac{\cup D E+\cup E A B}{2}=\frac{60^{\circ}+180^{\circ}}{2}=120^{\circ}
\end{gathered}
$$

Let $C K$ be the height of triangle $D C B$. Then

$$
C K=D K \operatorname{tg} \angle B D C=\frac{1}{2}
$$

Therefore,

$$
S_{\triangle \mathrm{DCB}}=\frac{1}{2} B D \cdot C K=\frac{\sqrt{3}}{4}
$$

Moreover,

$$
S_{\triangle \mathrm{ABE}}=\frac{1}{2} A B \cdot A E=1, S_{\triangle \mathrm{BDE}}=\frac{1}{2} B D \cdot D E=\frac{\sqrt{3}}{2}
$$

Therefore,

$$
S_{\mathrm{ABCDE}}=S_{\triangle \mathrm{ABE}}+S_{\triangle \mathrm{BDE}}+S_{\Delta \mathrm{DCB}}=1+\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{4}=1+\frac{3 \sqrt{3}}{4}
$$

## Answer

$1+\frac{3 \sqrt{3}}{4}$.

In triangle $A B C$, the height $B K$ and the median $M B$ are drawn to side $A C$, and $A M=B M$. Find the cosine of angle $K B M$, if $A B=1, B C=2$.

## Hint

Prove that the given triangle is a right triangle.

## Solution

According to problem $\underline{53377}$, triangle $A B C$ is a right triangle. We can reason in different ways from here.

First method. $A C^{2}=A B^{2}+B C^{2}=5, B K=A B \cdot B C / A C={ }^{2} / A C, \cos \angle K B M=B K / B M=4 / A C^{2}=4 / 5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-48.jpg?height=341&width=717&top_left_y=687&top_left_x=676)

Second method. $A K: K C=A B^{2}: B C^{2}=1: 4, \cos ^{2} \angle K B M={ }^{B K^{2}} / B M^{2}=4^{B K^{2}} / A C^{2}=4^{1 \cdot 4} /(1+4)^{2}=16 / 25$.

## Answer

$4 / 5$