Question

qual e a soma dos angulos internos de um pentagono?

Answer 1

$\textsf{Leia abaixo}$

Explicação passo a passo:

fórmula:

$\Large\boxed{\begin{array}{l} \Large\boxed{\begin{array}{l} \sf{S_i = (n - 2) \: . \: 180 {}^{ \circ} \rightarrow \begin{cases} \sf \: S_i = somar \,dos \,\hat{a}ngulos\, internos \\ \sf \: n =n\acute{u}mero\,de\, lados \end{cases}}\end{array}}\end{array}}$

Cálculos:

$\Large\boxed{\begin{array}{l} \Large\boxed{\begin{array}{l} \sf \:S_i = (n - 2) \: . \: 180 {}^{ \circ} \\ \\ \sf \:S_i = (5 - 2) \: . \: 180 {}^{ \circ} \\ \\ \sf \:S_i = 3 \: . \: 180 {}^{ \circ} \\ \\ \red{\boxed{ \boxed{ \sf{ \: \therefore \: S_i = 540 {}^{ \circ} \: }}}}\end{array}} \end{array}}$

Answer 2

$\large\boxed{\begin{array}{l} \rm{S_i = (n - 2) \cdot180 {}^{ \circ} } \\ \rm{S_i = (5 - 2) \cdot180 {}^{ \circ} } \\ \rm{S_i = 3 \cdot180 {}^{ \circ} } \\ \huge \boxed{ \boxed{ \boxed{ \boxed{ \rm{S_i = 540 {}^{ \circ} }}}}}\end{array}}$