10.230. The height, base, and sum of the lateral sides of the triangle are 24, 28, and 56 cm, respectively. Find the lateral sides.
Solution.

The area of the triangle $S=\frac{24 \cdot 28}{2}=336 \mathrm{~cm}^{2}$. Let one of the lateral sides be $x$ cm, then the second one is $(56-x)$ cm. The semi-perimeter of the triangle

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0735.jpg?height=444&width=680&top_left_y=77&top_left_x=338)

Fig. 10.39

$p=\frac{28+56}{2}=42$ (cm). Then, by Heron's formula, the area of the triangle $S=\sqrt{42 \cdot 14(42-x)(42-56+x)}$. Therefore, $\sqrt{3 \cdot 14^{2} \cdot(42-x)(x-14)}=336$; $\sqrt{3(42-x)(x-14)}=24 ;\left[\begin{array}{l}x=26, \\ x=30 .\end{array}\right.$ Thus, the lateral sides of the given triangle are 26 cm and 30 cm.

Answer: 26 cm; 30 cm.