Question

O perímetro do triângulo ABC da figura é:

Answer 1

Olá!
 
Apliquemos o Teorema de Pitágoras no triângulo ABD afim de determinar BD. Segue,

$\\ \displaystyle \mathsf{(5)^2 = \overline{BD}^2 + (2,2)^2} \\\\ \mathsf{\overline{BD}^2 = 25 - 4,84} \\\\ \mathsf{\overline{BD}^2 = 20,16} \\\\ \mathsf{\overline{BD} = \sqrt{20,16}}$


 Por conseguinte, aplicamos o Teorema de Pitágoras no triângulo BCD. Daí,

$\\ \displaystyle \mathsf{x^2 = \overline{BD}^2 + (10 - 2,2)^2} \\\\ \mathsf{x^2 = (\sqrt{20,16})^2 + (7,8)^2} \\\\ \mathsf{x^2 = 20,16 + 60,84} \\\\ \mathsf{x^2 = 81} \\\\ \mathsf{x = 9}$


 Por fim, encontramos o perímetro.

$\\ \displaystyle \mathsf{2p = \overline{AB} + \overline{BC} + \overline{CA}} \\\\ \mathsf{2p = 5 + 9 + 10} \\\\ \boxed{\mathsf{2p = 24}}$