aplicando as relações métricas nos triângulos retângulos abaixo, determine o valor:
(+20 pontos)
Anexos:
![](https://pt-static.z-dn.net/files/d98/4fc95327ef8018c0ebc213665bc69fb6.png)
Soluções para a tarefa
Respondido por
3
a) ![h^2=m.n \\
6^2=12\times n \\
36/12 = n\\
n= 3 h^2=m.n \\
6^2=12\times n \\
36/12 = n\\
n= 3](https://tex.z-dn.net/?f=h%5E2%3Dm.n++%5C%5C%0A6%5E2%3D12%5Ctimes+n+%5C%5C%0A36%2F12+%3D+n%5C%5C%0An%3D+3)
b)
c)![(2 \sqrt{6} )^2=3^2+y^2 \\
4\times6=9+y^2\\
24-9=y^2\\
\sqrt{15}=y (2 \sqrt{6} )^2=3^2+y^2 \\
4\times6=9+y^2\\
24-9=y^2\\
\sqrt{15}=y](https://tex.z-dn.net/?f=%282+%5Csqrt%7B6%7D+%29%5E2%3D3%5E2%2By%5E2+%5C%5C%0A4%5Ctimes6%3D9%2By%5E2%5C%5C%0A24-9%3Dy%5E2%5C%5C%0A+%5Csqrt%7B15%7D%3Dy+)
![(2 \sqrt{6})^2=x \times 3 \\
4\times6 = 3x \\
24= 3x\\
x= 8 (2 \sqrt{6})^2=x \times 3 \\
4\times6 = 3x \\
24= 3x\\
x= 8](https://tex.z-dn.net/?f=%282+%5Csqrt%7B6%7D%29%5E2%3Dx+%5Ctimes+3+%5C%5C%0A4%5Ctimes6+%3D+3x+%5C%5C%0A24%3D+3x%5C%5C%0Ax%3D+8+)
d)![h^2=m.n \\
h^2=2\times4 \\
h^2=8 \\
h=2 \sqrt{2} \\
h^2=m.n \\
h^2=2\times4 \\
h^2=8 \\
h=2 \sqrt{2} \\](https://tex.z-dn.net/?f=h%5E2%3Dm.n+%5C%5C%0Ah%5E2%3D2%5Ctimes4+%5C%5C%0Ah%5E2%3D8+%5C%5C%0Ah%3D2+%5Csqrt%7B2%7D+%5C%5C%0A)
![c^2=h^2+2^2 \\
c^2= \sqrt{8}^2+4 \\
c^2=8+4 \\
c= \sqrt{12} \\
c= 2\sqrt{3} c^2=h^2+2^2 \\
c^2= \sqrt{8}^2+4 \\
c^2=8+4 \\
c= \sqrt{12} \\
c= 2\sqrt{3}](https://tex.z-dn.net/?f=c%5E2%3Dh%5E2%2B2%5E2+%5C%5C%0Ac%5E2%3D++%5Csqrt%7B8%7D%5E2%2B4+%5C%5C%0Ac%5E2%3D8%2B4+%5C%5C%0Ac%3D++%5Csqrt%7B12%7D+%5C%5C%0Ac%3D+2%5Csqrt%7B3%7D++)
![a^2=b^2+c^2 \\
6^2=b^2+ \sqrt{12}^2 \\
36=b^2+12 \\
24=b^2 \\
b= 2 \sqrt{6} a^2=b^2+c^2 \\
6^2=b^2+ \sqrt{12}^2 \\
36=b^2+12 \\
24=b^2 \\
b= 2 \sqrt{6}](https://tex.z-dn.net/?f=a%5E2%3Db%5E2%2Bc%5E2+%5C%5C%0A6%5E2%3Db%5E2%2B+%5Csqrt%7B12%7D%5E2+%5C%5C%0A36%3Db%5E2%2B12+%5C%5C%0A24%3Db%5E2+%5C%5C%0Ab%3D+2+%5Csqrt%7B6%7D++)
b)
c)
d)
alfa1667:
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