Question

Qual o Numero de lados de um Poligono que possuem 14 diagonais ?

Answer 1

$D=\frac{n(n-3)}{2}\\\\14=\frac{n^2-3n}{2}\\\\14\cdot2=n^2-3n\\n^2-3x-28=0$

Coeficientes: a = 1, b = -3 e c = -28

$\Delta=b^2-4ac\\\Delta=(-3)^2-4\cdot1\cdot(-28)\\\Delta=9+112\\\Delta=121$

$n=\frac{-b+-\sqrt{\Delta}}{2a}\\\\n=\frac{-(-3)+-\sqrt{121}}{2\cdot1}\\\\n=\frac{3+-11}{2}\\\\n_1=\frac{3+11}{2}=\frac{14}{2}=7\\\\n_2=\frac{3-11}{2}=\frac{-8}{2}=-4$

Resposta: 7 lados

Conferindo:

$14=\frac{7(7-3)}{2}\\\\14=\frac{7\cdot4}{2}\\\\14=\frac{28}{2}\\\\14=14$



Espero ter ajudado.
Bons estudos! :)

Answer 2

d= n ( n - 3 )/2
14=(n2 -3n)/2
n2-3n=14x2
n2-3n=28
n2-3n-28=0
a=1
b=-3
c =-28
delta= (-3)² - 4.1.-28 
delta=9+112=121 raiz quadrada=11
x1 =( - ( -3)+11)/2
x1=(3+11):2   14/2=7 resposta o polígono é o heptágono 7 lados regulares