1.5. The base of the triangle is $30 \mathrm{~cm}$, and the lateral sides are 26 and $28 \mathrm{~cm}$. The height is divided in the ratio $2: 3$ (measured from the vertex), and a line parallel to the base is drawn through the point of division. Determine the area of the resulting trapezoid.
1.5. Given $A B=c=30 \mathrm{~cm}, A C=b=26 \mathrm{~cm}, B C=a=28 \mathrm{~cm}$ (Fig. 1.4). Then $p=0.5(a+b+c)=42, p-a=14, p-b=16, p-c=12$ and using Heron's formula, we find

$$
S_{\triangle A B C}=\sqrt{42 \cdot 14 \cdot 16 \cdot 12}=336\left(\mathrm{~cm}^{2}\right)
$$

Since $\triangle M N C \sim \triangle A B C$, then

$\frac{S_{\triangle M N C}}{S_{\triangle A B C}}=\frac{C K^{2}}{C D^{2}}=\frac{4}{25}=0.16$.

From this, we determine the area of the trapezoid:

$$
\begin{gathered}
S_{A M N B}=S_{\triangle A B C}-S_{\triangle M N C}= \\
=336-0.16 \cdot 336=282.24\left(\mathrm{~cm}^{2}\right) \\
\text { Answer: } 282.24 \mathrm{~cm}^{2}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-416.jpg?height=407&width=466&top_left_y=108&top_left_x=686)

Fig. 1.4