Question

Qual é o polígono que tem 104 diagonais? Resposta passo a passo.

Answer 1

$D=\frac{n(n-3)}{2}\\\\104=\frac{n^2-3n}{2}\\\\104\cdot2=n^2-3n\\208=n^2-3n\\n^2-3n-208=0$

Coeficientes: a = 1, b = -3 e c = -208

$\Delta=b^2-4ac\\\Delta=(-3)^2-4\cdot1\cdot(-208)\\\Delta=9+832\\\Delta=841$

$x=\frac{-b+-\sqrt{\Delta}}{2a}\\\\x=\frac{-(-3)+-\sqrt{841}}{2\cdot1}\\\\x=\frac{3+-29}{2}\\\\x_1=\frac{3+29}{2}=\frac{32}{2}=16\\\\x_2=\frac{3-29}{2}=\frac{-26}{2}=-13$


Resposta: hexadecágono



Espero ter ajudado.
Bons estudos! :)

Answer 2

Resolução da questão, vejamos:

$\faxmachine$

Vamos substituir isso nesta fórmula:

$\mathsf{D=\dfrac{n~\cdot~(n-3)}{2}}}}$

Observe:

$\mathsf{D=\dfrac{n~\cdot~(n-3)}{2}}}\\\\\\\\ \mathsf{104=\dfrac{n~\cdot~(n-3)}{2}}}}\\\\\\\\ \mathsf{208=n^{2}-3n}}}\\\\\\\\ \mathsf{n^{2}-3n-208=0}}}$

Agora vamos resolver essa equação do segundo grau, observemos:

$\mathsf{n=-b~\pm~\dfrac{\sqrt{b^{2}-4~\cdot~a~\cdot~c}}{2~\cdot~a}}}}~~\to\nathsf{n^{2}-3n-208=0}}\\\\\\\\ \mathsf{n=3~\pm~\dfrac{\sqrt{(-3)^{2}-4~\cdot~1~\cdot~(-208)}}{2~\cdot~1}}}}\\\\\\\\ \mathsf{n=3~\pm~\dfrac{\sqrt{9+832}}{2}}\\\\\\\\ \mathsf{n=3~\pm~\dfrac{\sqrt{841}}{2}}}\\\\\\\\ \mathsf{n'=\dfrac{3+29}{2}}}~=>~\Large\boxed{\boxed{\boxe{\mathsf{n'=16}}}}}}}}}\\\\\\\\ \mathsf{n''=\dfrac{3-29}{2}}}~=>\Large\boxed{\boxed{\mathsf{n"=-13.}}}}}}}}$

Ou seja, esse polígono é o Hexadecágono.

Espero que te ajude (^.^)