1.41. The ratio of the magnitudes of two angles of a triangle is 2, and the difference in the lengths of the opposite sides is 2 cm; the length of the third side of the triangle is 5 cm. Calculate the area of the triangle.
1.41. Given that $\angle A=2 \angle B, B C-A C=2$ cm, $A B=5$ cm (Fig. 1.33); 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, $\angle B=\angle D A C)$ and therefore $\frac{A C}{B C}=\frac{C D}{A C}$, i.e.

$$
A C^{2}=B C \cdot C D
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-432.jpg?height=382&width=356&top_left_y=505&top_left_x=799)

Fig. 1.33

Further, since $A D$ is the angle bisector, then

$$
\frac{A C}{C D}=\frac{5}{B D}, \text { or } \frac{A C}{C D}=\frac{5}{B C-C D} \text {, or } A C \cdot B C-A C \cdot C D=5 C D \text {, }
$$

from which

$$
C D=\frac{A C \cdot B C}{A C+5}
$$

From the equalities (1) and (2) it follows that $A C^{2}=\frac{A C \cdot B C^{2}}{A C+5}$, i.e. $A C^{2}+$ $+5 A C=A C^{2}+4 A C+4$, from which $A C=4 \mathrm{~cm}, B C=6 \mathrm{~cm}$. Thus,

$$
S_{\triangle A B C}=\sqrt{7.5 \cdot 1.5 \cdot 2.5 \cdot 3.5}=\frac{15 \sqrt{7}}{4}\left(\mathrm{~cm}^{2}\right)
$$

Answer: $3.75 \sqrt{7} \mathrm{~cm}^{2}$.