a diagonal de um quadrado mede 5 raiz de 2 cm. Determine a medida do lado
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Soluções para a tarefa
Respondido por
65
Um quadrado tem os seus lados iguais, vou chamar de x
Com a diagonal formamos um triangulo retângulo, onde
hipotenusa = 5√2
catetos = x
Teorema de Pitágoras
a²=b²+c²
(5√2)²=x²+x²
50=2x²
x²=50/2
x²=25
x=√25
x=5
O lado desse quadrado mede 5cm
Com a diagonal formamos um triangulo retângulo, onde
hipotenusa = 5√2
catetos = x
Teorema de Pitágoras
a²=b²+c²
(5√2)²=x²+x²
50=2x²
x²=50/2
x²=25
x=√25
x=5
O lado desse quadrado mede 5cm
Respondido por
24
Bom dia Luisa!
Solução!
![d=5 \sqrt{2}cm\\\\\
l=?\\\\\\
So~~aplicar~~o~~o~~teorema ~~de ~~ Pitagoras.](https://tex.z-dn.net/?f=d%3D5+%5Csqrt%7B2%7Dcm%5C%5C%5C%5C%5C%0Al%3D%3F%5C%5C%5C%5C%5C%5C%0ASo%7E%7Eaplicar%7E%7Eo%7E%7Eo%7E%7Eteorema+%7E%7Ede+%7E%7E+Pitagoras.%0A%0A "d=5 \sqrt{2}cm\\\\\
l=?\\\\\\
So~~aplicar~~o~~o~~teorema ~~de ~~ Pitagoras.")
![l^{2} +l^{2}=(5 \sqrt{2})^{2}\\\\\\
l^{2} +l^{2}=25.2\\\\\\
2l^{2}=50\\\\\\
l^{2}= \frac{50}{2}\\\\\\\
l^{2}= 25\\\\\\\
l= \sqrt{25} \\\\\\
l=5cm](https://tex.z-dn.net/?f=l%5E%7B2%7D+%2Bl%5E%7B2%7D%3D%285+%5Csqrt%7B2%7D%29%5E%7B2%7D%5C%5C%5C%5C%5C%5C%0Al%5E%7B2%7D+%2Bl%5E%7B2%7D%3D25.2%5C%5C%5C%5C%5C%5C%0A2l%5E%7B2%7D%3D50%5C%5C%5C%5C%5C%5C%0Al%5E%7B2%7D%3D+%5Cfrac%7B50%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%0Al%5E%7B2%7D%3D+25%5C%5C%5C%5C%5C%5C%5C+%0Al%3D+%5Csqrt%7B25%7D+%5C%5C%5C%5C%5C%5C%0Al%3D5cm+++++ "l^{2} +l^{2}=(5 \sqrt{2})^{2}\\\\\\
l^{2} +l^{2}=25.2\\\\\\
2l^{2}=50\\\\\\
l^{2}= \frac{50}{2}\\\\\\\
l^{2}= 25\\\\\\\
l= \sqrt{25} \\\\\\
l=5cm")
Bom dia!
Bons estudos!
Solução!
Bom dia!
Bons estudos!
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