10.161. A perpendicular is drawn to the hypotenuse through the vertex of the right angle of a right-angled triangle with legs of 6 and 8 cm. Calculate the areas of the resulting triangles.
Solution.

Let $C K$ be the perpendicular to the hypotenuse $A B$ of $\triangle A B C$ (Fig. 10.135).

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-631.jpg?height=494&width=306&top_left_y=81&top_left_x=255)

Fig. 10.135

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-631.jpg?height=492&width=474&top_left_y=82&top_left_x=671)

Fig. 10.136

Then $\quad A B=\sqrt{B C^{2}+A C^{2}}=10 \quad(\mathrm{~cm}) ; \quad S_{A B C}=\frac{1}{2} A C \cdot B C=24 \quad\left(\mathrm{~cm}^{2}\right)$; $\triangle B K C \sim \triangle A C B$. Therefore, $\frac{S_{A C B}}{S_{B K C}}=\left(\frac{A B}{B C}\right)^{2} \cdot$ Let's find

$$
S_{B K C}=\frac{S_{A C B} \cdot B C^{2}}{A B^{2}}=\frac{24 \cdot 64}{100}=15.36\left(\mathrm{~cm}^{2}\right) .
$$

Then $S_{A C K}=S_{A B C}-S_{B K C}=24-15.36=8.64\left(\mathrm{~cm}^{2}\right)$.

Answer: 15.36; 8.64$\left(\mathrm{~cm}^{2}\right)$.