Alguém me ajuda nessa questão ??????
Anexos:
Soluções para a tarefa
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Boa tarde!
Solução!
Para resolver esse exercício vamos usar semelhança de triângulos.
![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}](https://tex.z-dn.net/?f=Fazendo%21%5C%5C%5C%5C%5C%0A+%5Cdfrac%7B12%7D%7B13-y%7D+%3D+%5Cdfrac%7B13%7D%7B8%7D%5C%5C%5C%5C%5C%0A13%5Ctimes13-13%5Ctimes+y%3D12%5Ctimes8+%5C%5C%5C%5C%5C%0A169-13y%3D96+%5C%5C%5C%5C%5C%0A169-96%3D13y+%5C%5C%5C%5C%5C%0A73%3D13y+%5C%5C%5C%5C%5C%0Ay%3D+%5Cdfrac%7B73%7D%7B13%7D+%0A%0A%0A "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}")
![\dfrac{13}{8}= \dfrac{5}{x}\\\\\
13x=5.8\\\\\
13x=40 \\\\\
x= \dfrac{40}{13}](https://tex.z-dn.net/?f=+%5Cdfrac%7B13%7D%7B8%7D%3D+%5Cdfrac%7B5%7D%7Bx%7D%5C%5C%5C%5C%5C%0A13x%3D5.8%5C%5C%5C%5C%5C%0A13x%3D40+%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B40%7D%7B13%7D+%0A++ "\dfrac{13}{8}= \dfrac{5}{x}\\\\\
13x=5.8\\\\\
13x=40 \\\\\
x= \dfrac{40}{13}")
Vamos indicar os lados quadrilátero ABCD
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](https://tex.z-dn.net/?f=P%3D5%2B4%2B+%5Cdfrac%7B40%7D%7B13%7D%2B+%5Cdfrac%7B73%7D%7B13%7D%5C%5C%5C%5C%5C%0A13P%3D65%2B52%2B40%2B73%5C%5C%5C%5C%5C%5C%0A13P%3D230%5C%5C%5C%5C%5C%5C%0AP%3D+%5Cdfrac%7B230%7D%7B13%7D%5C%5C%5C%5C%5C%5C%5C%0AP%3D17%2C69++++ "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!
Solução!
Para resolver esse exercício vamos usar semelhança de triângulos.
Vamos indicar os lados quadrilátero ABCD
Perímetro a soma de todos os lados.
Boa tarde!
Bons estudos!
albertrieben:
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Oi Albert! Não esta conseguindo visualizar ?
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Ja cliquei!
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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
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
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