10.315. The ratio of the magnitudes of two angles of a triangle is 2, and the difference in the lengths of the sides opposite to them is 2 cm; the length of the third side of the triangle is 5 cm. Calculate the area of the triangle.
Solution.

$\angle A=2 \angle B, B C-A C=2$ cm, $A B=5$ cm (Fig. 10.104); we need to find $S_{\triangle A B C}$. Draw the angle bisector $A D$; then $\triangle A B C \sim \triangle A D C (\angle C-$ common, $\angle B=\angle D A C) \Rightarrow \frac{A C}{B C}=\frac{C D}{A C}, A C^{2}=B C \cdot C D\left(^{*}\right)$. Since $A D$ is the angle bisector, $\frac{A C}{C D}=\frac{5}{B D} \Rightarrow \frac{A C}{C D}=\frac{5}{B C-C D} \Rightarrow A C \cdot B C-A C \cdot C D=5 C D \Rightarrow$ $\Rightarrow C D=\frac{A C \cdot B C}{A C+5}\left({ }^{* *}\right)$. From the equalities (*) and (**) it follows that $A C^{2}=\frac{A C \cdot B C^{2}}{A C+5}$, $A C^{2}+5 A C=B C^{2}$. Since $B C=A C+2$, then $A C^{2}+5 A C=A C^{2}+4 A C+4$, from which $A C=4 \mathrm{~cm}, B C=6 \mathrm{~cm}$.

We obtained $S_{\triangle A B C}=\sqrt{7.5 \cdot 1.5 \cdot 2.5 \cdot 3.5}=\frac{15 \sqrt{7}}{4} \mathrm{~cm}^{2}$.

Answer: $\frac{15 \sqrt{7}}{4} \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0789.jpg?height=535&width=664&top_left_y=19&top_left_x=96)

Fig. 10.105

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0789.jpg?height=600&width=511&top_left_y=13&top_left_x=750)

Fig. 10.106