qual o comprimento da diagonal??
Anexos:
Soluções para a tarefa
Respondido por
1
Bom dia Caroline!
Solução!
Perímetro do retângulo.
Areá do retângulo!
Vamos substituir a equação I na equação II.
![(6-y).y=8\\\\ (6y-y^{2})-8=0\\\\\
-y^{2}+6y-8=0\\\\\
y^{2}-6y+8=0](https://tex.z-dn.net/?f=%286-y%29.y%3D8%5C%5C%5C%5C+%286y-y%5E%7B2%7D%29-8%3D0%5C%5C%5C%5C%5C%0A-y%5E%7B2%7D%2B6y-8%3D0%5C%5C%5C%5C%5C%0Ay%5E%7B2%7D-6y%2B8%3D0+++ "(6-y).y=8\\\\ (6y-y^{2})-8=0\\\\\
-y^{2}+6y-8=0\\\\\
y^{2}-6y+8=0")
Vamos aplicar a formula de Bhaskara para encontramos os lados do retângulo.
![y= \dfrac{-b\pm \sqrt{b^{2}-4.a.c } }{2.a} \\\\\\\
y=\dfrac{-(-6)\pm \sqrt{(-6)^{2}-4.1.8 } }{2.1} \\\\\\\
y=\dfrac{6\pm \sqrt{36-32 } }{2} \\\\\\\
y=\dfrac{6\pm \sqrt{4 } }{2} \\\\\\\
y=\dfrac{6\pm 2}{2} \\\\\\\\
y_{1} = \dfrac{6+2}{2}= \dfrac{8}{2} =4\\\\\\\\
y_{2}= \dfrac{6-2}{2}= \dfrac{4}{2}=2](https://tex.z-dn.net/?f=y%3D+%5Cdfrac%7B-b%5Cpm+%5Csqrt%7Bb%5E%7B2%7D-4.a.c+%7D+%7D%7B2.a%7D+%5C%5C%5C%5C%5C%5C%5C%0Ay%3D%5Cdfrac%7B-%28-6%29%5Cpm+%5Csqrt%7B%28-6%29%5E%7B2%7D-4.1.8+%7D+%7D%7B2.1%7D+%5C%5C%5C%5C%5C%5C%5C%0Ay%3D%5Cdfrac%7B6%5Cpm+%5Csqrt%7B36-32+%7D+%7D%7B2%7D+%5C%5C%5C%5C%5C%5C%5C%0Ay%3D%5Cdfrac%7B6%5Cpm+%5Csqrt%7B4+%7D+%7D%7B2%7D+%5C%5C%5C%5C%5C%5C%5C%0Ay%3D%5Cdfrac%7B6%5Cpm+2%7D%7B2%7D+%5C%5C%5C%5C%5C%5C%5C%5C%0Ay_%7B1%7D+%3D+%5Cdfrac%7B6%2B2%7D%7B2%7D%3D+%5Cdfrac%7B8%7D%7B2%7D+%3D4%5C%5C%5C%5C%5C%5C%5C%5C%0A+y_%7B2%7D%3D+%5Cdfrac%7B6-2%7D%7B2%7D%3D+%5Cdfrac%7B4%7D%7B2%7D%3D2++++ "y= \dfrac{-b\pm \sqrt{b^{2}-4.a.c } }{2.a} \\\\\\\
y=\dfrac{-(-6)\pm \sqrt{(-6)^{2}-4.1.8 } }{2.1} \\\\\\\
y=\dfrac{6\pm \sqrt{36-32 } }{2} \\\\\\\
y=\dfrac{6\pm \sqrt{4 } }{2} \\\\\\\
y=\dfrac{6\pm 2}{2} \\\\\\\\
y_{1} = \dfrac{6+2}{2}= \dfrac{8}{2} =4\\\\\\\\
y_{2}= \dfrac{6-2}{2}= \dfrac{4}{2}=2")
![(Diagonal)^{2} =(Largura)^{2} +(Comprimento)^{2}\\\\\\\
(Diagonal)^{2} =(2)^{2}+(4)^{2}\\\\\\\
(Diagonal)^{2} =4+16\\\\\\
(Diagonal)^{2} =20\\\\\\\
Diagonal= \sqrt{20}\\\\\\](https://tex.z-dn.net/?f=%28Diagonal%29%5E%7B2%7D+%3D%28Largura%29%5E%7B2%7D+%2B%28Comprimento%29%5E%7B2%7D%5C%5C%5C%5C%5C%5C%5C%0A%28Diagonal%29%5E%7B2%7D+%3D%282%29%5E%7B2%7D%2B%284%29%5E%7B2%7D%5C%5C%5C%5C%5C%5C%5C%0A%28Diagonal%29%5E%7B2%7D+%3D4%2B16%5C%5C%5C%5C%5C%5C%0A%28Diagonal%29%5E%7B2%7D+%3D20%5C%5C%5C%5C%5C%5C%5C%0ADiagonal%3D+%5Csqrt%7B20%7D%5C%5C%5C%5C%5C%5C+%C2%A0 "(Diagonal)^{2} =(Largura)^{2} +(Comprimento)^{2}\\\\\\\
(Diagonal)^{2} =(2)^{2}+(4)^{2}\\\\\\\
(Diagonal)^{2} =4+16\\\\\\
(Diagonal)^{2} =20\\\\\\\
Diagonal= \sqrt{20}\\\\\\")
Resposta: Alternativa D
Bom dia!
Bons estudos!
Solução!
Perímetro do retângulo.
Areá do retângulo!
Vamos substituir a equação I na equação II.
Vamos aplicar a formula de Bhaskara para encontramos os lados do retângulo.
Resposta: Alternativa D
Bom dia!
Bons estudos!
Usuário anônimo:
Carolina esta ai?
A coisa mais chata que tem: é uma pessoa mal educada que não responde quando perguntada sobre algo,ainda mais quando estamos ajudando.
Perguntas interessantes
Ed. Física,
10 meses atrás
Biologia,
10 meses atrás
Matemática,
10 meses atrás
Saúde,
1 ano atrás
História,
1 ano atrás
Matemática,
1 ano atrás