10.328. The sides of a triangle are 3, 4, and 5 cm. Determine the areas of the triangles into which the given triangle is divided by the height and median drawn to the largest side.
## Solution.

Let in $\triangle ABC$ (Fig. 10.116) $BC=3$ cm, $AC=4$ cm, $AB=5$ cm, $CK$ - median, $CM$ - altitude. Since $AB^2=BC^2+AC^2$, $\triangle ABC$ is a right triangle. Then $S_{\triangle ABC}=\frac{1}{2} BC \cdot AC=6 \text{ cm}^2$; $BM=\frac{BC^2}{AB}=\frac{9}{5} \text{ cm}$; $CM=\frac{BC \cdot AC}{AB}=\frac{12}{5} \text{ cm} \cdot S_{\triangle BMC}=\frac{1}{2} BM \cdot CM=\frac{54}{25}=2.16$ (cm²). Since $CK$ is a median, $S_{\triangle ACK}=S_{\triangle BCK}=\frac{1}{2} S_{\triangle ABC}=3 \text{ cm}^2. S_{\triangle CMK}=S_{\triangle BCK}-S_{\triangle BMC}=0.84 \text{ cm}^2$. Answer: $3 \text{ cm}^2; 0.84 \text{ cm}^2; 2.16 \text{ cm}^2$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0796.jpg?height=587&width=658&top_left_y=62&top_left_x=349)

Fig. 10.117