• Matéria: Matemática
  • Autor: nicolly241
  • Perguntado 8 anos atrás

determine x e y nas figuras:

Anexos:

Respostas

respondido por: FdASO
27
a)
y=m+n\\\\
m^2+1^2=(\sqrt{5})^2\\
m^2+1=5\\
m^2=4\\
m=\pm \ 2\\
m=2\\\\
y=2+n\\
n=y-2\\\\
n^2+1^2=x^2\\
n^2+1=x^2\\
(y-2)^2+1=x^2\\
y^2-4y+4+1=x^2\\
y^2-4y+5=x^2\\\\
y^2=(\sqrt{5})^2 + x^2\\
y^2=5+x^2\\\\
y^2-4y+5=x^2\\
5+x^2-4y+5=x^2\\
-4y=-10\\\\
y=\frac{-10}{-4}\\\\
y=\frac{5}{2}\\\\
y^2=5+x^2\\\\
(\frac{5}{2})^2=5+x^2\\\\
\frac{25}{4}=5+x^2\\\\
x^2=\frac{25}{4}-5\\\\
x^2=\frac{5}{4}\\\\
x=\frac{\sqrt{5}}{2}

b)

h^2+2^2=3^2\\
h^2+4=9\\
h^2=5\\
h=\sqrt{5}\\\\
h^2+x^2=y^2\\
(\sqrt{5})^2+x^2=y^2\\
5+x^2=y^2\\\\
3^2+y^2=(x+2)^2\\
9+y^2=x^2+4x+4\\
9+5+x^2=x^2+4x+4\\
4x=9+5-4\\
4x=10\\
x=\frac{5}{2}\\\\
y^2=x^2+5\\
y^2=(\frac{5}{2})^2+5\\\\
y^2=\frac{25}{4}+5\\\\
y^2=\frac{25+20}{4}\\\\
y^2=\frac{45}{4}\\\\
y=\sqrt{\frac{45}{4}}\\\\
y=\frac{3\sqrt{5}}{2}

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