Question

Alguém me ajuda nessa questão ??????

Answer 1

Boa tarde!

Solução!


Para resolver esse exercício vamos usar semelhança de triângulos.

$\triangle CDE \cong \triangle AEB$


$Fazendo!\\\\\ \dfrac{12}{13-y} = \dfrac{13}{8}\\\\\ 13\times13-13\times y=12\times8 \\\\\ 169-13y=96 \\\\\ 169-96=13y \\\\\ 73=13y \\\\\ y= \dfrac{73}{13}$

$\triangle AEB$

$\dfrac{13}{8}= \dfrac{5}{x}\\\\\ 13x=5.8\\\\\ 13x=40 \\\\\ x= \dfrac{40}{13}$


Vamos indicar os lados quadrilátero ABCD

$\overline{CD}=5\\\\\\ \overline{DC}=4\\\\\\ \overline{AB}= \dfrac{40}{13}\\\\\\ \overline{AD}= \dfrac{73}{13}$


Perímetro a soma de todos os lados.


$P=5+4+ \dfrac{40}{13}+ \dfrac{73}{13}\\\\\ 13P=65+52+40+73\\\\\\ 13P=230\\\\\\ P= \dfrac{230}{13}\\\\\\\ P=17,69$


Boa tarde!
Bons estudos!

Answer 2

Ola Etoo

tg(x) = 5/12

sen(x)/cos(x) = 5/12

12sen(x) = 5cos(x) 

cos(x) = 12sen(x)/5 

sen²(x) + cos²(x) = 1

sen²(x) + 144sen²(x)/25 = 1

169sen²(x) = 25

sen²(x) = 25/169

sen(x) = 5/13

cos(x) = 12sen(x)/5  = 12*5/(13*5) = 12/13 

sen(x) = AB/8 

5/13 = AB/8 

AB = 40/13 = 3.08 

AE² = 8² - (40/13)² = 9216/169
AE = √9216/13 = 7.38

AD = DE - AE = 13 - 7.38 = 5.62 

perimetro

P = AB + BC + CD + AD = 3.08 + 4 + 5 + 5.62 = 17.7