12.041. Given two sides $b$ and $c$ of a triangle and its area, equal to $0.4 b c$. Find the third side.
## Solution.

By the condition $S_{\triangle A B C}=0.4 \cdot b \cdot c$, where $b=A C, c=A B$ (Fig. 12.43). Let $C B=x$, then $S_{\triangle A B C}=\sqrt{p(p-c)(p-b)(p-x)}$, where $p=\frac{c+b+x}{2}$ (by Heron's formula) and $S_{\triangle A B C}=\sqrt{\frac{c+b+x}{2} \cdot \frac{b+x-c}{2} \cdot \frac{b+c-x}{2} \cdot \frac{x+c-b}{2}}=$ $=\frac{1}{4} \sqrt{4 b^{2} c^{2}-\left(b^{2}+c^{2}-x^{2}\right)^{2}}, 0.4 b c=\frac{\sqrt{4 b^{2} c^{2}-\left(b^{2}+c^{2}-x^{2}\right)^{2}}}{4} ;$ $2.56 b^{2} c^{2}=4 b^{2} c^{2}-\left(b^{4}+2 c^{2} b^{2}-2 x^{2} b^{2}+c^{4}-2 c^{2} x^{2}+x^{4}\right)$, from here $x^{4}-2 x^{2}\left(b^{2}+c^{2}\right)+b^{4}+c^{4}+0.56 b^{2} c^{2}=0, D=4 b^{4}+8 b^{2} c^{2}+4 c^{4}-4 b^{4}-4 c^{4}-$ $-2.24 b^{2} c^{2}=5.76 b^{2} c^{2}, x^{2}=\frac{2 b^{2}+2 c^{2} \pm 2.4 b \cdot c}{2} \Rightarrow x=\sqrt{b^{2}+c^{2} \pm 1.2 b c}$.

Answer: $\sqrt{b^{2}+c^{2} \pm 1.2 b c}$.

12..042. From a point taken on a circle of radius $R$, two equal chords are drawn, forming an inscribed angle equal to $\alpha$ radians.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-745.jpg?height=384&width=580&top_left_y=204&top_left_x=110)

Fig. 12.45

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-745.jpg?height=498&width=486&top_left_y=91&top_left_x=717)

Fig. 12.46

Find the part of the circle's area enclosed within this inscribed angle.

## Solution.

By the condition $A C=A B, \angle C A B=\alpha, R$ - radius of the circle (Fig. 12.44). $S_{C A B}=2 S_{\triangle C O A}+S_{C O B} ; S_{C O B}=\frac{1}{2} C O^{2} \cdot \angle C O B, \angle C O B=2 \angle C A B=2 \alpha$, then $S_{C O B}=\frac{1}{2} R^{2} \cdot 2 \alpha=R^{2} \cdot \alpha, \angle C O A=\frac{360^{\circ}-2 \alpha}{2}=180^{\circ}-\alpha$. We have that $S_{\triangle C O A}=\frac{1}{2} R^{2} \cdot \sin \left(180^{\circ}-\alpha\right)=\frac{R^{2}}{2} \sin \alpha ; S_{C A B}=R^{2} \cdot \sin \alpha+R^{2} \alpha=R^{2}(\alpha+\sin \alpha)$.

Answer: $R^{2}(\alpha+\sin \alpha)$.