23 DP-?

DEHP - прямоугольник. В прямоугольном треугольнике DHP, $$DP^2 = DH^2 + HP^2$$. $$DH = \sqrt{8}$$. $$HP = \sqrt{8} an(30^ ext{o}) = \sqrt{8} \cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{8}}{\sqrt{3}} = \frac{2\sqrt{2}}{\sqrt{3}} = \frac{2\sqrt{6}}{3}$$. $$DP^2 = (\sqrt{8})^2 + (\frac{2\sqrt{6}}{3})^2 = 8 + \frac{4 \cdot 6}{9} = 8 + \frac{24}{9} = 8 + \frac{8}{3} = \frac{24+8}{3} = \frac{32}{3}$$. $$DP = \sqrt{\frac{32}{3}} = \frac{4\sqrt{2}}{\sqrt{3}} = \frac{4\sqrt{6}}{3}$$.
Ответ: $$\frac{4\sqrt{6}}{3}$$