Question

encontre o numero de diagonais de um polígono convexo regular:
octógono
eneágono

Answer 1

fica assim:
octógono 
$\frac{8(8-3)}{2} = \frac{64-24}{2} = \frac{40}{2} d=20$
20 diagonais
eneágono
$\frac{9(9-3)}{2} = \frac{81-27}{2} = \frac{54}{2} d=27$
27 diagonais
espero ter ajudado!

Answer 2

fórmula:

$\blue{\Large\boxed{\begin{array}{l} \Large\boxed{\begin{array}{l} \sf{ d = \dfrac{n \cdot(n - 3)}{2} }\end{array}}\end{array}}}$

Cálculos:

octógono: 8 lados

$\Large\boxed{\begin{array}{l} \Large\boxed{\begin{array}{l} \sf{d = \dfrac{n \cdot(n - 3)}{2} } \\ \\ \sf{d = \dfrac{8 \cdot(8 - 3)}{2} } \\ \\ \sf{d = \dfrac{8 \cdot5}{2} } \\ \\ \sf{d = \dfrac{40}{2} } \\ \\ \red{ \boxed{ \boxed{ \sf{ \: \therefore \: d = 20}}}}\end{array}}\end{array}}$

eneágono: 9 lados

$\Large\boxed{\begin{array}{l} \Large\boxed{\begin{array}{l} \sf{d = \dfrac{n \cdot(n - 3)}{2} } \\ \\ \sf{d = \dfrac{9 \cdot(9 - 3)}{2} } \\ \\ \sf{d = \dfrac{9 \cdot6}{2} } \\ \\ \sf{d = \dfrac{54}{2} } \\ \\ \red{\boxed{ \boxed{ \sf{ \: \therefore \: d = 27}}}}\end{array}} \end{array}}$