Question

Observe atentamente a figura abaixo: *
O valor da incógnita x é?
x=15 x=40 x=25 x=29 x=35​

Answer 1

$\boxed{\begin{array}{l}\underline{\rm Pelo~teorema~de~Pit\acute agoras~temos:}\\\sf x^2+(x+1)^2=29^2\\\sf x^2+x^2+2x+1=841\\\sf 2x^2+2x+1-841=0\\\sf 2x^2+2x-840=0\div2\\\sf x^2+x-420=0\end{array}}$

$\boxed{\begin{array}{l}\sf\Delta=b^2-4ac\\\sf\Delta=1^2-4\cdot1\cdot(-420)\\\sf\Delta=1+1680\\\sf\Delta=1681\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf x=\dfrac{-1\pm\sqrt{1681}}{2\cdot1}\\\sf x=\dfrac{-1\pm41}{2}\begin{cases}\sf x_1=\dfrac{-1+41}{2}=\dfrac{40}{2}=20\\\sf x_2=\dfrac{-1-41}{2}=-\dfrac{42}{2}=-21\end{cases}\\\sf como~x>0,a~resposta~\acute e~x=20\\\sf corrija~o~gabarito\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm Demonstrac_{\!\!,}\tilde ao~da~resposta:}\\\sf os~catetos~medem~21~e~20~respectivamente.\\\sf a~hipotenusa~vale~29.\\\sf pelo~teorema~de~Pit\acute agoras,o~quadrado~da~hipotenusa\\\sf deve~ser~igual~a~soma~dos~quadrados~dos~catetos.\\\sf portanto\\\sf 21^2+20^2=441+400=841\\\sf note~que~29^2=29\cdot29=841\\\sf portanto~29^2=21^2+20^2\\\sf o~que~prova~que~o~valor~de~x~\acute e~20.\end{array}}$