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AC = √(29^2 - 20^2) = 21 cm

AB = √(35^2 - 21^2) = 28 cm

DB = 28 - 20 = 8 cm

Perimetro DBC = 8 + 29 + 35 = 72 cm

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Segmento $\overline{DB}:$

$\small\overline{DB}=\sqrt{\overline{BC}^2-\left(\sqrt{\overline{CD}^2- \overline{AD}^2}\right)^2}-\overline{AD}$

$\small \overline{DB}= \sqrt{35^2-\left(\sqrt{29^2-20^2}\right)^2}-20$

$\small \overline{DB}= \sqrt{1225-\left(\sqrt{841-400}\right)^2}-20$

$\small \overline{DB}= \sqrt{1225-\left(\sqrt{441}\right)^2}-20$

$\small \overline{DB}= \sqrt{1225-441}-20$

$\small \overline{DB}= \sqrt{784}-20$

$\small \overline{DB}= 28-20 = 8\,cm$

quindi:

perimetro del triangolo DBC $\small 2p_{(DBC)}= \overline{DB}+\overline{CD}+\overline{BC}= 8+29+35 = 72\,cm.$