1.32. Find the diagonal and the lateral side of an isosceles trapezoid with bases 20 and $12 \mathrm{~cm}$, given that the center of the circumscribed circle lies on the larger base of the trapezoid.
1.32. Since $A D$ is the diameter of the circle (Fig. 1.28), then $O D = O C = 10 \, \text{cm}$. Draw $C L \perp A D$; then $O L = 6$ cm and from $\triangle C L O$ we find $A$ $C L = \sqrt{O C^{2} - O L^{2}} = 8 \, (\text{cm})$. Now from $\triangle A L C$ and $\triangle C L D$ we get

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-428.jpg?height=561&width=602&top_left_y=110&top_left_x=549)

$A C = \sqrt{C L^{2} + A L^{2}} = \sqrt{64 + 256} = 8 \sqrt{5} \, (\text{cm})$,

Fig. 1.28

$C D = \sqrt{C L^{2} + L D^{2}} = \sqrt{64 + 16} = 4 \sqrt{5} \, (\text{cm})$.

Answer: $8 \sqrt{5}$ and $4 \sqrt{5}$ cm.