10.374 The bases of an isosceles trapezoid are 4 and 8 cm, and its area is 21 cm². Which side does the bisector of the angle at the larger base intersect: the smaller base or the lateral side of the trapezoid?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-231.jpg?height=243&width=438&top_left_y=298&top_left_x=161)

Let $C K$ be the height of the given isosceles trapezoid $A B C D, B C=4, A D=8$. Then

$$
\begin{aligned}
& K D=\frac{A D-B C}{2}=2, C K=\frac{2 S_{A B C D}}{4+8}=\frac{7}{2} \\
& C D=\sqrt{C K^{2}+K D^{2}}=\frac{\sqrt{67}}{2} \Rightarrow \\
& \Rightarrow A B=C D>B C \Rightarrow \angle C A D=\angle B C A>\angle B A C \Rightarrow
\end{aligned}
$$

$\Rightarrow$ the angle bisector of $\angle A$ will lie inside $\angle C A D$, i.e., it will intersect side $C D$.

Answer: the lateral side.