Problem 39. The height drawn to the base of an isosceles triangle is equal to 5 cm. The height drawn to the lateral side is 6 cm. Find the base.
Solution. Let $A D=5$ cm and $C K=6 \mathrm{~cm}$

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-033.jpg?height=477&width=534&top_left_y=567&top_left_x=361)
(Fig. 30). By formula II $a^{2}=2 b m$, but

$$
\begin{aligned}
& b=\sqrt{25+\frac{a^{2}}{4}}(\text { from } \triangle A D C) ; \\
& \left.m=\sqrt{a^{2}-36} \text { (from } \triangle C K B\right) .
\end{aligned}
$$

Then $a^{2}=2 \cdot \frac{\sqrt{100+a^{2}}}{2} \cdot \sqrt{a^{2}-36}$, or

$$
a^{4}=\left(100+a^{2}\right)\left(a^{2}-36\right)
$$

Expanding the brackets, we get:

$$
64 a^{2}=3600, a=7.5 \text { (cm) }
$$